Subject: Optional Mathematics

General equation of the second degree :

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0.

**General equation of second degree**

The general equation of first degree in x and y always represents a straight line.

Let A_{1}x + B_{1}y + C_{1} = 0.......................(i)

and A_{2}x + B_{2}y + C_{2} = 0.......................(ii)

be the equations of two straight lines.

Now, Combining these equations we get,

(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0........................(iii)

The coordinates of any point, which satisfy the equation (iii), will also satisfy either equation (i) or equation (ii).

Similarly, the co-ordinates of any point, which satisfy any one of the equation (i) or (ii) will also satisfy the equation (iii).

Therefore, equation (iii) represents two separate straight lines (i) and (ii). In other words, equation (iii) represents a pair of the straight lines given by (i) and (ii). So, (iii) is the equation of a pair of lines.

Now, Expanding the left hand side of equation (iii) we get,

A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0

If we put A_{1}A_{2} = a,A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1 }= 2f and C_{1}C_{2} = c, then the above equation becomes,

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0..........................(iv)

This equation is called the general equation of the second degree in x and y. Thus we see that the equation of a pair of lines is a second degree equation. But the converse of this statement is not always true. It means every second degree equations in x and y may not represent a pair of straight lines. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.

Consider an equation y^{2} - 3xy + 2x^{2} = 0. This equation is equivalent to (y - x) (y - 2x) = 0.

So, the equation y^{2} - 3xy + 2x^{2} = 0 represents the two straight lines y - x = 0 and y - 2x = 0.

Similarly, the equation xy = 0 represents the two straight lines x = 0 and y = 0.

Again,

Consider an equation x^{2} - 5x + 6 = 0 represents two straight lines x - 2 = 0 and x - 3 = 0.

And the equation x^{2} - y^{2} = 0 represents the two straight lines x + y = 0 and x - y = 0.

**Condition that the general equation of second degree may represent a line pair**

The general equation of second degree in x and y is

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

or, ax^{2} + (2hy + 2g)x + (by^{2} + 2fy + c) = 0

This is quadratic equation in x.

So, x = \(\frac {-(2hy + 2g) ± \sqrt {(2hy + 2g)^2 - 4a(by^2 + 2fy + c)}}{2a}\)

or, x = \(\frac {-(hy + g) ± \sqrt {(hy + g)^2 - a(by^2 + 2fy + c)}}{a}\)

These two equations will be linear if

(hy + g)^{2} - a (by^{2} + 2fy + c) is a perfect square.

or, h^{2}y^{2} + 2ghy + g^{2} - aby^{2} - 2afy - ac is a perfect square.

i.e. (h^{2} - ab) y^{2} + (2gh - 2af) y + (g^{2} - ac) is a perfect square.

i.e. (2gh - 2af)^{2} - 4 (h^{2} - ab) (g^{2} - ac) = 0

i.e. g^{2}h^{2} - 2ghaf + a^{2}f^{2} - g^{2}h^{2} + h^{2}ac + abg^{2} - a^{2}bc = 0

i.e. a (af^{2} + bg^{2} + ch^{2} - 2fgh - abc) = 0

i.e. abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0

Hence, the general equation of second degree ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 may represent a line pair if abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0.

**Quadratic Equation**

Any equation in the form of ax^{2} + bx + c = 0 is called quadratic equation in x.

Multiplying both sides of this equation by 4a we get,

4a (ax^{2} + bx + c) = 0

or, 4a^{2}x^{2} + 4abx + 4ac = 0

or, (2ax)^{2} + 2 . 2ax . b + b^{2} - b^{2} + 4ac = 0

or, (2ax + b)^{2} = b^{2} - 4ac

or, (2ax + b)^{2} = (\(\sqrt {b^2 - 4ac}\))^{2}

or, 2ax + b = ± \(\sqrt {b^2 - 4ac}\)

or, 2ax = - b ± \(\sqrt {b^2 - 4ac}\)

or, x = \(\frac {- b ± \sqrt {b^2 - 4ac}}{2a}\)

Let α = \(\frac {- b + \sqrt {b^2 - 4ac}}{2a}\) and β = \(\frac {- b - \sqrt {b^2 - 4ac}}{2a}\).

Then α and β are called roots of the quadratic equation ax^{2} + bx + c = 0.

Now,

\begin{align*} \text{Sum of the roots (α + β)} &= \frac {- b + \sqrt {b^2 - 4ac}}{2a} + \frac {- b - \sqrt {b^2 - 4ac}}{2a}\\ &= - \frac ba\\ &= -\frac {coefficient\;of\;x}{coefficient\;of\;x^2}\\ \end{align*}

\begin{align*} \text{Sum of the roots (αβ)} &= (\frac {- b + \sqrt {b^2 - 4ac}}{2a}) (\frac {- b - \sqrt {b^2 - 4ac}}{2a})\\ &= \frac ca\\ &= \frac {constant\;term}{coefficient\;of\;x^2}\\ \end{align*}

**Homogeneous equation of second degree**

An equation in x and y in which the sum of the power of x and y in every term is the same, is called homogenous equation. If this sum is two, then the equation is called a homogenous equation of second degree. The equation ax^{2} + 2hxy + by^{2} = 0 is the general homogenous equation of the second degree.

**A homogeneous equation of the second degree represents a pair of straight lines which pass through the origin.**

**Proof: **Consider the homogenous equation of the second degree.

ax^{2 }+ 2hxy + by^{2} = 0...........................(i)

or, by^{2} + 2hxy + ax^{2} = 0

If b ≠ 0, the equation can be written as y^{2} + \(\frac {2h}b\)xy + \(\frac ab\)x^{2} = 0

or, (\(\frac yx\))^{2} + \(\frac {2h}b\)(\(\frac yx\)) + \(\frac ab\) = 0

This is quadratic equation in \(\frac yx\). So it has two roots. Let these roots be m_{1} and m_{2}.

Then,

\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}

or, y = m_{1}x and y = m_{2}x.

These two equations are the equations of straight lines passing through the origin.

If b = 0, then equation (i) becomes

ax^{2} + 2hxy = 0

or, x(ax + 2hy) = 0 which represents two straight lines x = 0 and ax + 2hy = 0.

These two lines pass through the origin.

Hence, the homogenous equation of second degree always represents a pair of straight lines passing through the origin.

**Angle between the line pair represented by ax ^{2} + 2hxy + by^{2} = 0**

Homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0

or, by^{2} + 2hxy + ax^{2} = 0

or, y^{2} + \(\frac{2h}{b}\)(\(\frac yx\)) + \(\frac ab\) = 0.........................(i)

This is quadratic equation in \(\frac yx\). So, it has two roots. Let these two roots be m_{1} and m_{2}.

Then,

\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}

or, y = m_{1}x and y = m_{2}x which are two seperate equations represente by the given equation ax^{2} + 2hxy + by^{2} = 0.

Now,

From the quadratic equation (i)

m_{1} + m_{2} = \(\frac {-2h}b\) and m_{1}m_{2} = \(\frac ab\)

Now,

\begin{align*} m_1- m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac{4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x. Then,

\begin{align*} tan\theta &= ± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

∴ \(\theta\) = tan^{-1 }(± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

**Second Method:**

We have ax^{2} + 2hxy + by^{2} = 0............................................(i)

Let the seperate equations represented by this equation be y = m_{1}x and y = m_{2}x.

Now,

The combined equation of theses equation is:

(y - m_{1}x) (y - m_{2}x) = 0

or, y^{2} - (m_{1} + m_{2})xy + m_{1}m_{2}x^{2} = 0

or, m_{1}m_{2}x^{2}- (m_{1} + m_{2})xy + y^{2} = 0.......................................(ii)

Comparing (i) and (ii) we have,

\(\frac {m_1m_2}{a}\) = \(\frac {- (m_1 + m_2)}{2h}\) = \(\frac 1b\)

Taking 1^{st} and last, m_{1}m_{2} = \(\frac ab\)

Taking 2^{nd} and last, m_{1} + m_{2} = -\(\frac {2h}b\)

Now,

\begin{align*} m_1 - m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac {4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x.

Then,

\begin{align*} tan\theta &=± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

∴ \(\theta\) = tan^{-1} (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

**Condition that the straight lines given by the equation ax ^{2} + 2hxy + by^{2} = 0 may be (1) Perpendicular and (2) Coincident.**

- If a + b = 0 the value of tan\(\theta\) is∞ and hence \(\theta\) is 90°. Hence two straight lines represented by the equation ax
^{2}+ 2hxy + by^{2}= 0 are perpendicular to each other if**coefficient of x**^{2}+ coefficient of y^{2}= 0.

For example: the equations x^{2}- y^{2}= 0 and 6x^{2}+ 11xy - 6y^{2}= 0 both represent pairs of straight lines at right angles.

Similarly, whatever be the value of h, the equation x^{2}+ 2hxy - y^{2}= 0 represents a pair of straight lines at right angles. - If h
^{2}= ab, the value of tan\(\theta\) is zero and hence \(\theta\) is zero. But both the straight lines represented by ax^{2}+ 2hxy + by^{2}= 0 pass through the origin. So, the two lines are coincident.

Hence, two straight lines represented by the equation ax^{2}+ 2hxy + by^{2}= 0 are coincident if**h**This may be seen directly from the original equation ax^{2}= ab.^{2}+ 2hxy + by^{2}= 0. If h^{2}= ab i.e. h = \(\sqrt {ab}\), then the original equation becomes,

ax^{2}+ 2\(\sqrt {ab}\)xy + by^{2}= 0

or, (\(\sqrt a\)x)^{2}+ 2 \(\sqrt a\) \(\sqrt b\) xy + (\(\sqrt b\)y)^{2}= 0

or, (\(\sqrt a\)x + \(\sqrt b\)y)^{2}= 0 which gives two coincident straight lines

i.e.\(\sqrt a\)x + \(\sqrt b\)y = 0 and\(\sqrt a\)x + \(\sqrt b\)y = 0.

**If the equation ax ^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines, then ax^{2} + 2hxy + by^{2} = 0 represents a pair of lines through the origin parallel to the above pair.**

**Proof:**

Let ax^{2}+ 2hxy + by^{2} + 2gx + 2fy + c = 0............................................................(i)

represent a pair of straight lines. Then the left-hand side can be resolved into two linear factors. Let these factors be A_{1}x + B_{1}y + C_{1}and A_{2}x + B_{2}y + C_{2} = 0.

Now

Combining equation of these equations is:

(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0

or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0..............................................(ii)

Equating the coefficients of like terms in equation (i) and (ii) we have,

A_{1}A_{2} = a^{2}, A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b^{2}, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1} = 2f, C_{1}C_{2} = c^{2}

Now,

Equation of the straight line parallel to A_{1}x + B_{1}y + C_{1} = 0 and passing through the origin is:

A_{1}x+ B_{1}y = 0.........................................(iii)

Again,

Equation of straight line parallel to A_{2}x + B_{2}y + C_{2} = 0 and passing through the origin is:

A_{2}x + B_{2}y = 0.........................................(iv)

Now,

Combining (iii) and (iv) we have,

(A_{1}x + B_{1}y) (A_{2}x + B_{2}y) = 0

or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} = 0

or, ax^{2} + 2hxy + by^{2} = 0. This completes the proof.

**Note:**Angles between the line pair ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 are same as the angles between the pair ax^{2} + 2hxy + by^{2} = 0.

The angles are given by,

tan^{-1} (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

Hence, the two lines will be perpendicular to each other if a + b = 0 and they will beparallel if h^{2}= ab.

General equation of the second degree :

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Find the equation of the lines represented by:

3x^{2} - 5xy - 2y^{2} - x + 2y = 0

Here,

3x^{2} - 5xy - 2y^{2} - x + 2y = 0

or, 3x^{2} - 6xy + xy - 2y^{2} - x + 2y = 0

or, 3x(x - 2y) + y(x - 2y) - 1(x - 2y) = 0

or, (x - 2y) (3x + y - 1) = 0

Either: x - 2y = 0

Or: 3x + y - 1 = 0

∴ The required equations are: x - 2y = 0 and 3x + y - 1 = 0_{Ans}

Find the separate equation of the line given by equation:

x^{2} - 3xy + 2y^{2} = 0

Here,

x^{2} - 3xy + 2y^{2} = 0

or, x^{2} - 2xy - xy + 2y^{2} = 0

or, x(x - 2y) - y(x - 2y) = 0

or, (x - 2y) (x - y) = 0

Either: x - 2y = 0

Or: x - y = 0

∴ The required seperate equations are:x - 2y = 0 andx - y = 0 _{Ans}

Show that the straight lines represented by the equation 6x^{2} + 11xy - 6y^{2} = 0 are perpendicular to each other.

Here,

The given equation is:6x^{2} + 11xy - 6y^{2} = 0..................(1)

The homogeneous equation of second is: ax^{2} + 2hxy + by^{2} = 0.................(2)

Comparing (1) and (2)

a = 6

2h = 11 i.e. h = \(\frac {11}2\)

b = -6

Now.

a + b = 0

or, 6 - 6 = 0

∴ 0 = 0

Hence, a + b = 0 is satisfied by the given equation so the equation6x^{2} + 11xy - 6y^{2} = 0 are perpendicular to each other. _{Proved}

Prove that the straight lines represented by the equation x^{2} - 4xy + 4y^{2} = 0 are coincident to each other.

Here,

Given equation is:x^{2} - 4xy + 4y^{2} = 0......................(1)

The homogenous equation of second degree is:

ax^{2}+ 2hxy + by^{2} = 0...........................(2)

Comparing (1) and (2)

a = 1

2h = -4 i.e. h = \(\frac {-4}2\) = -2

b = 4

Now,

h^{2} = ab

or, (-2)^{2} = 1× 4

∴ 4 = 4

Hence, h^{2} = ab is satisfied by the given equation so the equationx^{2} - 4xy + 4y^{2} = 0 are coincident to each other. _{Proved}

Write down the condition for the lines represented by the equation ax^{2} + 2hxy + by^{2} = 0 to be coincident and perpendicular to each other.

Here,

ax^{2} + 2hxy + by^{2} = 0

When: h^{2} = ab then straight lines are coincident each other.

When: a + b = 0 then straight lines are perpendicular each other.

Find the homogeneous equation of the second degree from the pair of lines:

x = 2y and 2x = y

Given eq^{n} of line are:

x = 2y

i.e. x - 2y = 0..........................(1)

2x = y

i.e. 2x - y = 0..........................(2)

Combined eq^{n} of (1) and (2) is:

(x - 2y) (2x - y) = 0

or, 2x^{2}- xy - 4xy + 2y^{2} = 0

or, 2x^{2} - 5xy + 2y^{2} = 0

∴ The required eq^{n} is: 2x^{2} - 5xy + 2y^{2} = 0 _{Ans}

Find a pair of eq^{n} of straight lines represented by y^{2} = x^{2} and hence prove that they are perpendicular to each other.

Here,

Given eq^{n} is: y^{2} = x^{2}

or, x^{2} - y^{2} = 0

or, (x - y) (x + y) = 0

either: x + y = 0

Or, x - y = 0

Slope of eq^{n} x + y = 0 is: m_{1} = -\(\frac 11\) = -1

Slope of eq^{n} x -y = 0 is: m_{2} = -\(\frac 1{-1}\) = 1

m_{1}× m_{2} = -1× 1 = -1

Hence, they are perpendicular to each other. _{Proved}

Prove that the angles between the lines represented by 6x^{2} - 5xy - 6y^{2} = 0 is at right angle.

Here,

Given eq^{n} is:

6x^{2} - 5xy - 6y^{2} = 0...............................(1)

The eq^{n}of homogenous is:

ax^{2} + 2hxy + by^{2} = 0..........................(2)

Comparing eq^{n} (1) and (2)

a = 6

b = -6

Now,

a + b = 0

or, 6 - 6 = 0

∴ 0 = 0

Hence, the angle between two lines is 90°. _{Proved}

Find the angle between a pair of straight lines represented by the equation x^{2} - 2xy sec\(\alpha\) + y^{2} = 0.

Here,

Given eq^{n} is:

x^{2} - 2xy sec\(\alpha\) + y^{2} = 0...........................(1)

The homogenous eq^{n} is:

ax^{2} + 2hxy + by^{2} = 0..............................(2)

Comparing eq^{n} (1) and (2)

a = 1

h = - sec\(\alpha\)

b = 1

Now,

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {(-sec\alpha)^2 - 1 × 1}}{1 + 1}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {sec^2\alpha - 1}}2\)

or, tan\(\theta\) =± \(\sqrt {sec^2\alpha - 1}\)

or, tan\(\theta\) =± \(\sqrt {tan^2\alpha}\)

or, tan\(\theta\) =± tan\(\alpha\)

∴ \(\theta\) = \(\alpha\) _{Ans}

Find a single equations representing the line pairs x cos\(\alpha\) +y sin\(\alpha\) = 0 and x sin\(\alpha\) + y cos\(\alpha\) = 0.

Here,

Given eq^{n} are:

x cos\(\alpha\) + y sin\(\alpha\) = 0...................(1)

x sin\(\alpha\) + y cos\(\alpha\) = 0...................(2)

Combined eq^{n} of (1) and (2)

(x cos\(\alpha\) + y sin\(\alpha\)) (x sin\(\alpha\) + y cos\(\alpha\)) = 0

or, x^{2} cos\(\alpha\).sin\(\alpha\) +xy cos^{2}\(\alpha\) + xy sin^{2}\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) = 0

or, x^{2} cos\(\alpha\).sin\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) + xy (cos^{2}\(\alpha\) + sin^{2}\(\alpha\)) = 0

or, sin\(\alpha\).cos\(\alpha\) (x^{2} + y^{2}) + xy× 1 = 0

∴(x^{2} + y^{2}) sin\(\alpha\).cos\(\alpha\) + xy = 0 _{Ans}

Find the obtuse angle between a pair of lines represented by the equation:

x^{2} + 4xy + y^{2} = 0

Given equation is:

x^{2} + 4xy + y^{2} = 0.............................(1)

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a = 1

h =2

b = 1

Now,

tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {2^2 - 1 × 1}}{1 + 1}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {4 - 1}}2\)

or, tan\(\theta\) = ± \(\sqrt 3\)

∴ obtuse angle (\(\theta\) = tan^{-1} (-\(\sqrt 3\))

∴ \(\theta\) = (90 + 30)° = 120°_{Ans}

Find the angle between two straight lines represented by:

3x^{2} + 2y^{2} - 5xy = 0

Here,

Given equation is:

3x^{2} + 2y^{2} - 5xy = 0

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a = 1

h =2

b = 1

Now,

tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = \(\frac {2\sqrt {(\frac {-5}2)^2 - 3 × 2}}{3 + 2}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25}4 - \frac 61}}5\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25 - 24}4}}5\)

or, tan\(\theta\) = \(\frac {2 × \frac 12}{\frac 51}\)

or, tan\(\theta\) = \(\frac 15\)

or, \(\theta\) = tan^{-1}(\(\frac 15\))

∴ \(\theta\) = 11.31° _{Ans}

Find the obtuse angle between the lines represented by the equation 12x^{2} - 23xy + 5y^{2} = 0.

Here,

The given equation is:

12x^{2} - 23xy + 5y^{2} = 0.....................(1)

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a =12

h = -\(\frac {23}2\)

b = 5

Now,

tan\(\theta\) = ±\(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-23}2)^2 - 12 × 5}}{12 + 5}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529}4 - 60}}{17}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529 - 240}4}}{17}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {289}}2\)× \(\frac 1{17}\)

or,tan\(\theta\) = ± \(\frac {17}{17}\)

∴tan\(\theta\) = ± 1

Taking +ve sign,

tan\(\theta\) = 1

or, tan\(\theta\) = tan 45°

∴ \(\theta\) = 45°

Taking -ve sign,

tan\(\theta\) = -1

or, tan\(\theta\) = tan (180 - 35)

or, tan\(\theta\) = tan 135°

∴ \(\theta\) = 135°

Hence, the obtuse angle between the pair of equation is: 135°. _{Ans}

Find the acute angle between the pair of lines through origin represented by homogeneous equation of the second degree 2x^{2} - 5xy + 2y^{2} = 0.

Here,

The homogenous equation of second degree is:

ax^{2} + 2hxy +by^{2} = 0................................(1)

The given equation is:

2x^{2} - 5xy + 2y^{2} = 0...................................(2)

Comparing (1) and (2)

a = 2

2h = -5 i.e. h = -\(\frac 52\)

b = 2

If \(\theta\) be the angle between pair of lines:

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {(\frac {-5}2)^2 - 2 × 2}}{2 + 2}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - \frac 41}}4\)

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {25 - 16}4}}2\)

or, tan\(\theta\) =± \(\sqrt {\frac 94}\)× \(\frac 12\)

or, tan\(\theta\) =± \(\frac 32\)× \(\frac 12\)

or, tan\(\theta\) =± \(\frac 34\)

∴ \(\theta\) = tan^{-1}(± \(\frac 34\))

Hence, the acute angle (\(\theta\)) =tan^{-1}(± \(\frac 34\)) _{Ans}

Find the separate equation of the lines contained by the equation 6x^{2} + 5xy - 3x - 2y - 6y^{2} = 0 and prove that two lines are perpendicular each other.

Here,

Given equation is:

6x^{2} + 5xy - 3x - 2y - 6y^{2} = 0

or, 6x^{2} + 9xy - 4xy - 6y^{2} - 3x - 2y = 0

or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0

or, (2x + 3y) (3x - 2y - 1) = 0

Either: 2x + 3y = 0..........................(1)

Or: 3x - 2y - 1 = 0............................(2)

Slope of equation (1), m_{1} = -\(\frac 23\)

Slope of equation (2), m_{2} = \(\frac 32\)

Again,

m_{1}× m_{2} = \(\frac {-2}3\)× \(\frac 32\) = -1

The product of two slopes = -1

Hence, these equations are perpendicular each other. _{Hence, Proved}

To find the equation of a pair of straight lines through origin.

Here,

Let: the two equation of the lines through origin be:

a_{1}x + b_{1}y = 0..............................(1)

a_{2}x + b_{2}y = 0..............................(2)

Combined equation of (1) and (2) is:

(a_{1}x + b_{1}y) (a_{2}x + b_{2}y) = 0

or, a_{1}a_{2}x^{2} + a_{1}b_{2}xy + a_{2}b_{1}xy + b_{1}b_{2}y^{2} = 0

or,a_{1}a_{2}x^{2} + (a_{1}b_{2}+ a_{2}b_{1})_{}xy + b_{1}b_{2}y^{2} = 0

Let:

a_{1}a_{2} = a

b_{1}b_{2} = b

(a_{1}b_{2}+ a_{2}b_{1}) = 2h

Now,

ax^{2} + 2hxy + by^{2} = 0

Thus,ax^{2} + 2hxy + by^{2} = 0 is homogenous equation of second degree. _{Ans}

To prove that the homogenous equations of second degree ax^{2} + 2hxy + by^{2} = 0 always represents a pair of straight lines through origin.

Here,

The given equation is: ax^{2} + 2hxy + by^{2} = 0 if a≠ 0, then:

The given equation is multiplied by 'a' on both sides:

a^{2}x^{2} + 2ahxy + aby^{2} = 0

or (ax)^{2} + 2 (ax) (hy) + (hy)^{2} - h^{2}y^{2} + aby^{2} = 0

or, (ax + hy)^{2} - (h^{2} - ab) y^{2} = 0

or, (ax + hy)^{2} - {(\(\sqrt {(h^2 - ab)y})}^{2} = 0

or, (ax + hy + \(\sqrt {(h^2 - ab)y}\)) (ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0

Either:(ax + hy + \(\sqrt {(h^2 - ab)y}\)) = 0.................(1)

Or:(ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0..........................(2)

Equation (1) and (2) are satisfied (0, 0) so both straight lines will passes through origin.

Again,

If a = 0

The equation ax^{2}+ 2hxy + by^{2} = 0 will be

2hxy + by^{2} = 0

or, y(2hx + by) = 0

Either: y = 0.......................(3)

Or: 2hx + by = 0..............(4)

Equation (3) and (4) are satisfied by the co-ordinates (0, 0).

Hence, the second degree homogeneous equation: ax^{2} + 2hxy + by^{2} = 0 always represents a pair of straight lines through the origin. _{Proved}

Find the angles between the pair of lines represented by the equation ax^{2} + 2hxy + by^{2} = 0.

Here,

Given equationax^{2} + 2hxy + by^{2}= 0 represents two straight lines that passes through the origin.

Let: y = m_{1}x and y = m_{2}x are the two lines.

y -m_{1}x = 0............................(1)

y - m_{2}x = 0............................(2)

Product of the equation (1) and (2) is:

(y - m_{1}x) (y - m_{2}x) = 0

or, m_{1}m_{2}x^{2} - m_{1}xy - m_{2}xy + y^{2} = 0

or, m_{1}m_{2}x^{2} - (m_{1 }+m_{2})xy + y^{2} = 0...................(3)

Given equationax^{2} + 2hxy + by^{2}= 0

\(\frac ab\)x^{2} + \(\frac {2h}b\) xy + y^{2} = 0...................(4)

Comparing equation (3) and (4), we get:

m_{1} + m_{2} = -\(\frac {2h}b\) and m_{1}m_{2} = \(\frac ab\)

Let \(\theta\) be the angle between the lines:

tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)

or, tan\(\theta\) =± \(\frac {\sqrt {(m_1 + m_2)^2 - 4m_1m_2}}{1 + m_1m_2}\) [\(\because\) a - b = \(\sqrt {(a + b)^2 - 4ab}\)]

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2}{b^2} - \frac {4a}b}}{1 + \frac ab}\)

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2 - 4ab}{b^2}}}{\frac {b + a}b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)× \(\frac bb\)

or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

∴\(\theta\) = tan^{-1}(± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))_{ Ans}

Find the single equation passing through the point (1, 1) and parallel to the lines represented by the equation x^{2} - 5xy + 4y^{2} = 0.

Here,

x^{2} - 5xy + 4y^{2} = 0

or, x^{2} - xy - 4xy + 4y^{2} = 0

or, x(x - y) - 4y(x - y) = 0

or, (x - y) (x - 4y) = 0

Either: x - y = 0.................(1)

Or: x - 4y = 0......................(2)

The eq^{n} (1) changes into parallel form is:

x - y + k_{1} = 0......................(3)

The point (1, 1) passes through eq^{n} (1)

1 - 1 + k_{1} = 0

∴ k_{1} = 0

Putting the value of k_{1} in eq^{n} (3)

x - y + 0 = 0

x - y = 0............................(4)

The eq^{n}(2) changes in parallel form is:

x - 4y + k_{2} = 0..................(5)

The point (1, 1) passes through eq^{n} (5)

1 - 4× 1 + k_{2} = 0

or, -3 + k_{2} = 0

∴ k_{2} = 3

Putting the value of k_{2} in eq^{n} (5)

x - 4y + 3 = 0.......................(6)

The eq^{n} of pair of line is:

(x - y) (x - 4y + 3) = 0

or, x^{2} - 4xy + 3x - xy - 4y^{2} - 3y = 0

∴x^{2} - 5xy - 4y^{2} + 3x - 3y = 0_{Ans}

Find the single equation of pair of straight lines passing through the origin and perpendicular to the line pairs represented by x^{2} - xy - 2y^{2} = 0.

Here,

Given equation is:

x^{2} - xy - 2y^{2} = 0

or, x^{2} - 2xy + xy - 2y^{2} = 0

or, x (x - 2y) + y (x - 2y) = 0

or, (x - 2y) (x + y) = 0

Either: x - 2y = 0.........................(1)

Or: x + y = 0.................................(2)

The eq^{n} (1) changes in perpendicular form is:

-2x - y + k_{1} = 0

2x + y - k_{1} = 0..........................(3)

The eq^{n} (3) passes through origin (0, 0):

2× 0 + 0 - k_{1} = 0

∴ k_{1} = 0

Putting the value of k_{1} in eq^{n} (3)

2x + y - 0 = 0

2x + y = 0........................(4)

The eq^{n} (2) change in perpendicular form is:

x - y + k_{2} = 0........................(5)

The eq^{n} (5) passes through origin (0, 0)

0 - 0 + k_{2} = 0

∴ k_{2} = 0

Putting the value of k_{2} in eq^{n} (5)

x - y + 0 = 0

x - y = 0...........................(6)

The equation of the pairs of lines is:

(2x + y) (x - y) = 0

or, 2x^{2} - 2xy + xy - y^{2} = 0

∴2x^{2} - xy - y^{2} = 0_{Ans}

Find the single equation of the pair of straight lines passing through the point (2, 3) and perpendicular to the line 3x^{2} - 8xy + 5y^{2}^{ }= 0.

Given eq^{n} of line is:

3x^{2} - 8xy + 5y^{2}^{}= 0

or, 3x^{2} - 3xy - 5xy + 5y^{2}^{}= 0

or, 3x (x - y) - 5y (x - y) = 0

or, (x - y) (3x - 5y) = 0

Either: x - y = 0.......................(1)

Or: 3x - 5y = 0.........................(2)

The eq^{n} (1) changes in perpendicular form is:

x + y + k_{1} = 0...........................(3)

The point (2, 3) passes through eq^{n} (3)

2 + 3 + k_{1} = 0

or, 5 + k_{1} = 0

∴ k_{1} = -5

Putting the value of k_{1} in eq^{n} (3)

x + y - 5 = 0..........................(4)

The eq^{n} (2) change in perpendicular form is:

5x + 3y + k_{2} = 0...................(5)

The point (2, 3) passes through eq^{n} (5)

5× 2 + 3× 3 + k_{2} = 0

or, 10 + 9 + k_{2} = 0

or, 19 + k_{2} = 0

∴ k_{2} = - 19

Putting the value of k_{2} in eq^{n} (5)

5x + 3y - 19 = 0......................(6)

The eq^{n} of pairs of lines is:

(x + y - 5) (5x + 3y - 19) = 0

or, 5x^{2} + 3xy - 19x + 5xy + 3y^{2} - 19y - 25x - 15y + 95 = 0

∴ 5x^{2} + 8xy + 3y^{2} - 44x - 34y + 95 = 0_{Ans}

Find the equation of two lines which passes through the point (3, -1) and perpendicular to the line pair x^{2} - xy - 2y^{2} = 0.

Here,

x^{2} - xy - 2y^{2} = 0

or, x^{2} - 2xy + xy - 2y^{2} = 0

or, x(x - 2y) + y(x - 2y) = 0

or, (x - 2y) (x + y) = 0

Either: x - 2y = 0..................(1)

Or: x + y = 0...........................(2)

The eq^{n} (1) change in perpendicular form is:

2x + y + k_{1} = 0.......................(3)

The point (3, -1) passes through eq^{n} (3)

2× 3 - 1 + k_{1} = 0

or, 6 - 1 + k_{1} = 0

∴ k_{1}= -5

Putting the value of k_{1} in eq^{n} (3)

2x + y - 5 = 0..........................(4)

The eq^{n} (2) change in perpendicular form is:

x - y + k_{2} = 0.........................(5)

The point (3, -1) passes througheq^{n} (5)

3 + 1 + k_{2} = 0

∴ k_{2} = -4

Putting the value of k_{2} in eq^{n} (5)

x - y - 4 = 0.........................(6)

The eq^{n} of pair of lines is:

(2x + y - 5) (x - y - 4) = 0

or, 2x^{2} - 2xy - 8x + xy - y^{2} - 4y - 5x + 5y + 20 = 0

∴ 2x^{2} - xy- y^{2} - 13x + y + 20 = 0_{Ans}

Find the equation of the pair of lines represented by the equation 2x^{2} + 5xy + 3y^{2} = 0. Also find the angle between them.

Here,

2x^{2} + 5xy + 3y^{2} = 0

or, 2x^{2} + 3xy + 2xy + 3y^{2} = 0

or, x(2x + 3y) + y(2x + 3y) = 0

or, (2x + 3y) (x + y) = 0

Either: 2x + 3y = 0

Or: x + y = 0

Given eq^{n} is: 2x^{2} + 5xy + 3y^{2} = 0......................(1)

Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0...................(2)

Comparing eq^{n} (1) and (2)

a = 2

h = \(\frac 52\)

b = 3

We know that:

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or,tan\(\theta\) =± \(\frac {2\sqrt {(\frac 52)^2 - 2 × 3}}{2 + 3}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - 6}}5\)

or,tan\(\theta\) =± \(\frac {2\sqrt {\frac {25 - 24}4}}5\)

or,tan\(\theta\) =± \(\frac {2\sqrt {\frac 14}}5\)

or,tan\(\theta\) =± \(\frac {2 × \frac 12}5\)

or,tan\(\theta\) =± \(\frac 15\)

Taking +ve sign,

\(\theta\) = tan^{-1} (\(\frac 15\)) = 11.31°

Taking -vesign,

\(\theta\) = (180° - 11.31°) = 168.69°

∴ Required eq^{n} are: 2x + 3y = 0 and x + y = 0 and angle between them are: 11.31° and 168.69°. _{Ans}

If \(\alpha\) be the angle made by the straight lines represented by the equation x^{2} + 2xy sec\(\theta\) + y^{2} = 0. Prove that: \(\alpha\) = \(\theta\). Find the pairs of equations of the lines.

Given eq^{n} is:x^{2} + 2xy sec\(\theta\) + y^{2} = 0........................(1)

Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0..........(2)

Comparingeq^{n}(1) and (2)

a = 1

h = sec\(\theta\)

b = 1

If \(\alpha\) be the angle between pairs of lines then,

tan\(\alpha\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1 × 1}}{1 + 1}\)

or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1}}2\)

or, tan\(\alpha\) = \(\sqrt {tan^2\theta}\)

or, tan\(\alpha\) = tan\(\theta\)

∴\(\alpha\) = \(\theta\) _{Proved}

Again,

The given eq^{n} is:

y^{2} + 2xy sec\(\theta\) + x^{2} = 0..........................(3)

The quadratic eq^{n} is:

ax^{2} + bx + c = 0.............................(4)

Comparing (3) and (4)

a = 1

b = 2x sec\(\theta\)

c = x^{2}

We know,

x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)

or, y = \(\frac {-(2x sec\theta) ± \sqrt {(2x sec\theta)^2 - 4 × 1 × x^2}}{2 × 1}\)

or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 sec^2\theta - 4x^2}}2\)

or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 (sec^2\theta - 1)}}2\)

or, y = \(\frac {-2x sec\theta ± 2x\sqrt {tan^2\theta}}2\)

or, y = \(\frac {-2x sec\theta ± 2x tan\theta}2\)

or, y = \(\frac {2(-x sec\theta ± xtan\theta)}2\)

or, y = - xsec\(\theta\)± x tan\(\theta\)

Taking +ve sign:

y = -xsec\(\theta\) + xtan\(\theta\)

Taking -ve sign:

y = -xsec\(\theta\) - xtan\(\theta\)

∴ The required eq^{n} are:y = -xsec\(\theta\) + xtan\(\theta\) andy = -xsec\(\theta\) - xtan\(\theta\) _{Ans}

Prove that the angle between the lines represented by the equation x^{2} - 2xy cosec\(\theta\) + y^{2} = 0 is (\(\frac p2\) - \(\theta\)).

Given equation is:x^{2} - 2xy cosec\(\theta\) + y^{2} = 0.....................(1)

Homogenous equation is: ax^{2} + 2hxy + by^{2} = 0......................(2)

Comparing eq^{n} (1) and (2)

a = 1

h = -cosec\(\theta\)

b = 1

If \(\alpha\) be the angle between pair of lines then:

tan\(\alpha\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1}}2\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)

or, tan\(\alpha\) =± \(\sqrt {cot^2\theta}\)

∴ tan\(\alpha\) =± cot\(\theta\)

Taking +ve sign,

tan\(\alpha\) = cot\(\theta\)

tan\(\alpha\) = tan(\(\frac p2\) - \(\theta\))

∴\(\alpha\) = (\(\frac p2\) - \(\theta\))_{Hence, Proved}

Find the equations of two lines represented by the equation 2x^{2} + 7xy + 3y^{2} = 0. Also find the angle between them.

Here,

Given equation is:

2x^{2} + 7xy + 3y^{2} = 0

or, 2x^{2} + 6xy + xy + 3y^{2} = 0

or, 2x(x + 3y) + y(x + 3y) = 0

or, (x + 3y)(2x + y) = 0

The two equation represented by2x^{2} + 7xy + 3y^{2} = 0 are:

2x + y = 0......................(1)

x + 3y = 0......................(2)

Now,

Slope of equation (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = -2

Slope of equation (2),m_{2} =-\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 13\)

If the angle between the lines be \(\theta\) then:

tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)

or, tan\(\theta\) =± \(\frac {-2 + \frac 13}{1 + (-2) (-\frac 13)}\)

or, tan\(\theta\) = ± \(\frac {\frac {-6 + 1}3}{\frac {3 + 2}3}\)

or, tan\(\theta\) = ± \(\frac {-5}3\)× \(\frac 35\)

or, tan\(\theta\) = ± 1

Taking +ve sign,

tan\(\theta\) = tan 45°

∴ \(\theta\) = 45°

Taking -ve sign,

tan\(\theta\) = tan (180 - 45)° = tan 135°

∴ \(\theta\) = 135°

∴ Required angles (\(\theta\)) = 45° and 135°_{Ans}

Find the equation of a pair of lines represented by the equation 2x^{2} + 3xy - 2y^{2} = 0. Also find the angle between them.

The given equation is:

2x^{2} + 3xy - 2y^{2} = 0

or, 2x^{2} + 4xy - xy - 2y^{2} = 0

or, 2x(x + 2y) - y(x + 2y) = 0

or, (x + 2y) (2x - y) = 0

∴ Equation are: x + 2y = 0 and 2x - y = 0

Again,

2x^{2} + 3xy - 2y^{2} = 0............................(1)

The homogenous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0........................(2)

Comparingequation (1) and (2)

a = 2

h = \(\frac 32\)

b = -2

tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = \(\frac {2\sqrt {(\frac 32)^2 - 2 × (-2)}}{2 + (-2)}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac 94 + \frac 41}}{2 - 2}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {9 + 16}4}}0\)

or, tan\(\theta\) =∞

∴ \(\theta\) = tan^{-1}∞ = 90°

∴ The required equations are: x + 2y = 0 and 2x - y = 0 and angle between the pairs of straight lines is 90°._{Ans}

Find the obtuse angle between the pair of lines represented by the equation 2x^{2} - 3xy + y^{2} = 0. Also find the pair of lines.

Here,

Given equation is:

2x^{2}- 3xy + y^{2} = 0......................(1)

The homogenous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0..................(2)

Comparing equation (1) and (2)

a = 2

h = -\(\frac 32\)

b =1

Now,

tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-3}2)^2 - 2 × 1}}{2 + 1}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac 94 - \frac 21}}3\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac {9 - 8}4}}3\)

or, tan\(\theta\) = ± \(\frac {2 × \frac 12}3\)

or, tan\(\theta\) = ± \(\frac 13\)

For the obtuse angle,

tan\(\theta\) = -\(\frac 13\)

\(\theta\) = tan^{-1} (\(\frac 13\))

Now,

2x^{2} - 3xy + y^{2} = 0

or, 2x^{2} - 2xy - xy + y^{2} = 0

or, 2x(x - y) - y(x - y) = 0

or, (x - y) (2x - y) = 0

∴ The required pairs of lines are: x - y = 0 and 2x - y = 0. _{Ans}

Find separate equation of the lines contained by the equation 6x^{2} + 5xy - 3x + 2y - 6y^{2} = 0 and show that the lines are perpendicular to each other.

Here,

6x^{2} + 5xy - 3x + 2y - 6y^{2} = 0

or, 6x^{2} + 5xy - 6y^{2} - 3x + 2y = 0

or, 6x^{2} + 9xy - 4xy - 6y^{2} - 3x + 2y = 0

or, 3x(2x + 3y) - 2y(2x + 3y) - 1(3x - 2y) = 0

or, (2x + 3y - 1) (3x - 2y) = 0

Either: 2x + 3y - 1 = 0...........(1)

Or: 3x - 2y = 0.........................(2)

Slope of equation (1), m_{1} = \(\frac {-3}{-2}\) = \(\frac 32\)

Slope of equation (2),m_{2} = -\(\frac 23\)

∴ m_{1}× m_{2} = \(\frac 32\)× \(\frac {-2}3\) = -1

Hence, the product of two slopes = -1 so these equations are perpendicular to each other. _{Proved}

Find the equation of the pair of lines represented by the equation x^{2} - 2xy cosec\(\theta\) + y^{2} = 0. Also find the angle between them.

The given equation is:

x^{2} - 2xy cosec\(\theta\) + y^{2} = 0

or, (y)^{2} - (2x cosec\(\theta\)) y + (x)^{2} = 0..................(1)

The quadratic equation is:

ax^{2} + bx + c = 0......................(2)

Comparing (1) and (2)

a = 1

b = -2x cosec\(\theta\)

c = x^{2}

x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)

y = -(-2x cosec\(\theta\))± \(\frac {\sqrt {(-2x cosec\theta)^2 - 4(1) (x^2)}}{2× 1}\)

y = \(\frac {2x cosec\theta ± \sqrt {4x^2 cosec^2\theta - 4x^2}}2\)

y = \(\frac {2x cosec\theta ± 2x \sqrt {cosec^2\theta - 1}}2\)

y = \(\frac {2(x cosec\theta ± x cot\theta)}2\)

y = x cosec\(\theta\)± x cot\(\theta\)

The pair of lines are:

y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and

y - x (cosec\(\theta\) + cot\(\theta\)) = 0

Comparing x^{2} -2xy cosec\(\theta\) + y^{2} = 0 and ax^{2} + 2hxy + by^{2} = 0

a = 1

h = - cosec\(\theta\)

b = 1

We know that:

tanA =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tanA = ± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)

or, tanA =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)

or, tanA =± cot\(\theta\)

∴ A = tan^{-1} (± cot\(\theta\))

∴ The required equations are: y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and y - x (cosec\(\theta\) + cot\(\theta\)) = 0 and angle (A) = tan^{-1} (± cot\(\theta\)). _{Ans}

© 2019-20 Kullabs. All Rights Reserved.