For the curve with and

MATHEMATICS · Single correct · CBSE · Mathematics · Conic Sections · Conic Sections

Problem

For the curve

y^2 = 4ax

with

a = 2

and

k \neq 0

, the point

(k, 3k)

lies on the curve when

k

equals: (a)

\dfrac{9}{8}

(b)

\dfrac{8}{9}

(c)

\dfrac{2}{3}

(d)

\dfrac{2}{9}