Question

Tangents are drawn from (-8 ,-14) to the parabola y^2 = 16x .Find their eqns and the pt of contact

Answer 1

Answer:

Tangent

x + 4y + 64 = 0

has point of contact at ( 64, -32 ).

Tangent

2x - y + 2 = 0

has point of contact at ( 1, 4 ).

Step-by-step explanation:

Let (x, y) be a point on the parabola such that the tangent there passes through (-8, -14).

The slope of the line is

( y + 14 ) / ( x + 8 )

Since it is a tangent to the parabola, its slope is also the value of dy / dx on the curve at (x, y).

y² = 16 x  =>  2y dy / dx = 16  =>  dy / dx  = 8 / y

Therefore

( y + 14 ) / ( x + 8 )  =  8 / y

=> y ( y + 14 ) = 8 ( x + 8 )

=> y² + 14y = 8x + 64

=> y² + 14y = y²/2 + 64   ( since 16x = y², so 8x = y²/2 )

=> y²/2 + 14y = 64

=> y² + 28y = 128

=> ( y + 14 )² = 128 + 14² = 128 + 196 = 324

=> y + 14 = ± 18

=> y = - 14 ± 18

=> y = -32 or 4.

When y = -32, we have x = y²/16 = 64

so the point of contact is ( 64, -32 ).

The gradient of the tangent is

dy/dx = 8 / y = 8 / -32 = - 1 / 4.

So the equation of the tangent is

( y + 14 ) / ( x + 8 ) = - 1 / 4

=> 4y + 56 = -x -8

=> x + 4y + 64 = 0

When y = 4, we have x = y²/16 = 1

so the point of contact is ( 1, 4 ).

The gradient of the tangent is

dy/dx = 8 / y = 8 / 4 = 2.

The equation of the tangent is

( y + 14 ) / ( x + 8 ) = 2

=> y + 14 = 2x + 16

=> 2x - y + 2 = 0