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