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

x² + (3x/2)² = 52

x² + (9x²/4) = 52

 13x²/4 = 52

x² = 16

x = ± 4

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

Example 1 :

Find the equations of normal to

y = x³ - 3x

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

Solution :

Slope of tangent drawn to the circle.

dy/dx = 3x² - 3(1)

= 3x² - 3

Slope of the normal line to the curve 

= -1/(3x² - 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/(3x² - 3)

3x² - 3 = 9

x² = 12/3

x² = 4

x = √4

x =  ±2

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

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