FIND EQUATION OF TANGENT TO THE CURVE WHICH IS PARALLEL TO THE LINE

Example 1 :

Find the equations of those tangents to the circle

x2 + y2 = 52

which are parallel to the straight line 2x + 3y = 6.

Solution :

If two lines are parallel, then slopes of those lines will be equal.

Here,

slope of tangent = slope of the given line

Slope of the given line 2x + 3y = 6

m = -coefficient of x/coefficient of y

m = -2/3 ----(1)

Slope of the tangent line drawn to the circle

2x + 2y(dy/dx) = 0

dy/dx = -x/y ----(2) 

(1) = (2)

-2/3 = -x/y

2y = 3x

y = 3x/2

Substituting the value of y in the equation of curve.

x2 + y2 = 52

x+ (3x/2)2 = 52

x+ (9x2/4) = 52

 13x2/4 = 52

x2 = 16

x = ± 4

If x = 4, then y = 6

If x = -4, then y = -6

So, the point of contact are (4, 6) (-4, -6).

Equation of the tangent passing through the point (4, 6) and m = -2/3.

y - y1 = m(x - x1)

y - 6 = (-2/3)(x - 4)

3(y - 6) = -2(x - 4)

3y - 18 = -2x + 8

2x + 3y - 8 - 18 = 0

2x + 3y - 26 = 0 

Equation of the tangent passing through the point (-4, -6) and m = -2/3.

y + 6 = (-2/3)(x + 4)

3(y + 6) = -2(x + 4)

3y + 18 = -2x - 8

2x + 3y + 26 = 0

Example 1 :

Find the equations of normal to

y = x- 3x

that is parallel to 2x + 18y - 9 = 0.

Solution :

Slope of tangent drawn to the circle.

dy/dx = 3x- 3(1)

= 3x- 3

Slope of the normal line to the curve 

= -1/(3x- 3) ----(2)

Slope of the line parallel to the normal line to the curve

m = -coefficient of x/coefficient of y

m = -2/18

m = -1/9 ----(1)

(1) = (2)

-1/9 = -1/(3x- 3)

3x- 3 = 9

x2 = 12/3

x= 4

x = √4

x =  ±2

x = 2

y = 2- 3(2)

y = 2

x = -2

y = (-2)- 3(-2)

y = -2

points of contact are (2, 2) (-2, -2).

Equation of the normal passing through the point

(2, 2) and m = -1/9

y - 2 = (-1/9)(x - 2)

9(y - 2) = -1(x - 2)

9y - 18 = -x + 2

x + 9y - 20 = 0

Equation of the normal passing through the point

(-2, -2) and m = -1/9

y + 2 = (-1/9)(x + 2)

9(y + 2) = -1(x + 2)

9y + 18 = -x - 2

 x + 9y + 20 = 0

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