Notes on Tangents and Normals to any curve | Grade 11 > Mathematics > Coordinate Geometry straight line | KULLABS.COM

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### Introduction:

#### Tangent:

In geometry, the tangent line (or simply tangent) to a plane curveat a givenpointis the straight linethat just touches the curve at that point.Leibniz defined it as the line through a pair of infinitely close points on the curve.More precisely, a straight line is said to be a tangent of a curve y = f (x) at a point x = c on the curve if the line passes through the point (c, f (c)) on the curve and has slope f '(c)where f ' is the derivative of f. A similar definition applies to space curvesand curves in n-dimensionalEucidian space.

As it passes through the point where the tangent line and the curve meet, called the point of tangency, the tangent line is "going in the same direction" as the curve, and is thus the best straight-line approximation to the curve at that point.

Similarly, the tangent plane to a surfaceat a given point is the planethat just touches the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometryand has been extensively generalized.

The word "tangent" comes from the Latin word tangere which mean'to touch'.

Importance of tangent

The "tangent point" is that point for which you want to calculate the slope. It might give you the instantaneous acceleration of an object at that point in a velocity-time graph. It is essentially that point at which you want to know what the slope is. And to calculate the slope, usually you would want the x and y coordinates of the "tangent point."

Characteristics of tangent

1. A tangent to a circle is perpendicular to the radius at the point of tangency.

2.The angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord.

3.A tangent is straight line and only touches the point of curve.

Normal

A normal is a line which is perpendicular to the tangent.

### Some theorems

Theorem 1

Tangent on any point on the circle which are as follows

$$(i)point(P)=(x_1,y_1)$$

$$(ii)circle=x^2+y^2=a^2$$

Here,

The given circle and points are respectively:-

$$(x_1,y_1)…(i)$$

$$x^2+y^2=a^2…(ii)$$

Since the given point lies on the given circle so,

$$(x_1)^2+(y_1)^2=a^2…(iii)$$

Now center of circle (iii) is O(0,0)

Now slope of line OP is:-

$$\frac{y_1-0}{x_1-0}$$

$$=\frac{y_1}{x_1}$$

Since, the tangent at the point P is perpendicular to the line OP. So, slope of the tangent at point P equals to

$$\frac{-x_1}{y_1}$$

(As the product of slopes of the perpendicular lines is -1)

Hence, equation of tangent at the point P is

$$y-y_1=frac{-x_1}{y_1}(x-x_1)$$

$$or, yy_1-y^2=-xx_1+(x_1)^2$$

$$or, xx_1+ yy_1=(x_1)^2+(y_1)^2$$

$$xx_1+ yy_1=a^2$$

Using equation (iii)

$$∴ xx_1+ yy_1=a^2$$

Is the required equation of tangent on the circle at any point.

Theorem 2

Tangent of the circle at the respective point and the circle.

$$Point=(x_1,y_1)$$

$$circle=x^2+y^2+2gx+2fy+c=0(General Form)$$

Here,

We can write Equation of the circle at a given point as:-

$$( x_1)^2+(y_1)^2+2g(x_1)+2f(y_1)+c=0….(i)$$

$$( x_1,y_1)$$

We can assume another Point

$$Q(x_2,y_2)$$

Such that we get equation of circle as

$$( x_2)^2+(y_2)^2+2g(x_2)+2f(y_2)+c=0…(ii)$$

Now, subtracting (i) from (ii) we get,

$$( x_2)^2-( x_1)^2+(y_2)^2-(y_1)^2+2g(x_2-x_1)+2f(y_2-y_1)=0$$

$$(x_2-x_1)(x_2+x_1+2g)+(y_2-y_1)(y_2+y_1+2f)=0$$

$$\frac{y_2-y_1}{x_2-x_1}=\frac{x_1+x_2+2g}{y_1+y_2+2f}…(iii)$$

Now the equation of secant PQ is

$$(y-y_1)=\frac{-2x_1+2g}{2y_1+2f}(x-x_1)$$

$$i.e (y-y_1)=\frac{-x_1+x_2+2g}{y_1+y_2+2f}(x-x_1)…….(iv) using(iii)$$

As Q approaches and concides with P, the secant PQ becomes the tangent of P

Hence equation (iv) becomes

$$(y-y_1)=\frac{-2x_1+2g}{2y_1+2f}(x-x_1)$$

$$or,(y-y_1)(y_1+f)+(x-x_1)(x_1+g)=0$$

$$or,xx_1+yy_1+gx+fy=x_1^2+y_1^2+gx_1+fy_1$$

Adding x1g+y1f+c to both side we get

$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=x_1^2+y_1^2+2x_1g+2y_1f+c$$

Using (i) the equation of tangent is:-

$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$$T

Thorem 3

Condition of straight line to be the tangent of the circle in the form:

$$x^2+y^2=a^2$$

Here,

we can assume equation of straight line as

$$y=mx+c$$

$$or,mx-y+c=0..(i)($$

Now,

Center of circle is (0,0)

And let a be the radius.

To be the straight line a tangent the perpendicular distance from tangent to center should be equal to the radius of circle.

$$i.e\frac{0.m+0+c}{1+m^2}=a$$

$$∴c=a\sqrt{1+m^2}$$

It is the required condition of tangency

Theorem 4

Equation of normal at given point and circle as:

$$point=(x_1,y_1)$$

$$circle=x^2+y^2=a^2$$

Here equation of tangent at the point P(x1,y1) is

$$xx_1+yy_1=a^2$$

And slope of tangent is

$$\frac{-x_1}{y_1}$$

As tangent is perpendicular to normal so

Slope of normal is

$$\frac{y_1}{x_1}$$

Hence the equation of normal at point p(x1,y1) is :-

$$y-y_1=\frac{y_1}{x_1}(x-x_1)$$

$$∴xy_1=yx_1$$

Which is the required equation

$$xx_1+ yy_1=a^2$$

$$( x_2)^2+(y_2)^2+2g(x_2)+2f(y_2)+c=0$$

$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$$

$$c=a\sqrt{1+m^2}$$

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