Let L be a tangent line to the parabola y^2=4x-20 at (6,2).

Question

Let $L$ be a tangent line to the parabola $y^2=4x-20$ at $(6,2)$. If $L$ is also a tangent to the ellipse $\dfrac{x^2}2+\dfrac{y^2}b=1$, then the value of $b$ is equal to

Accepted Answer

Parabola y2 = 4x - 20

Tangent at P(6, 2) will be

$$2y = 4\left( {{{x + 6} \over 2}} \right) - 20$$

2y = 2x + 12 - 20

2y = 2x - 8

y = x - 4

x - y - 4 = 0 ....... (1)

This is also tangent to ellipse $${{{x^2}} \over 2} + {{{y^2}} \over b} = 1$$

Apply c2 = a2m2 + b2

(-4)2 = (2)(1) + b

b = 14

Solutions · 1