Question

If three points E, F, G are taken on the parabola $y^2=4ax$ so that their ordinates are in G.P., then the tangents at E and G intersect on the

• A. directrix
• B. axis
• C. ordinate of F
• D. tangent at F

Answer 1

Answer: (C)

Given parabola is $y^2=4ax...\left(1\right)$
Let the coordinates of E, F and G be respectively
$\left(a{t}_1^2,2a{t}_1\right),\left(a{t}_2^2,2a{t}_2\right)$ and $\left(a{t}_3^2,2a{t}_3\right)$
Since ordinates of E, F and G are in G.P.
$\therefore {\left(2a{t}_2\right)}^2=\left(2a{t}_1\right)\left(2a{t}_3\right)$ or $t_2^2=t_1{t}_3...\left(2\right)$
The tangents at E and G are
$t_{1}y=x+a{t}_1^2...\left(3\right)$
and $t_{3}y=x+a{t}_3^2$
Solving $\left(3\right)$ and $\left(4\right)$, we get $x=a{t}_1{t}_3=a{t}_2^2$ [from $\left(2\right)$]
Since the x-coordinate of the point of intersection is $a{t}_2^2$, the point lies on the line $x=a{t}_2^2$ i.e. on the ordinate of $F\left(a{t}_2^2,2a{t}_2\right).$