Question

If the tangents to the parabola $y^2=4ax$ at $(x_1,y_1),(x_2,y_2)$ cut at $(x_3,y_3)$ then

• A. $x_1,x_3,x_2$ are in A.P.
• B. $x_1,x_3,x_2$ are in G.P.
• C. $y_1,y_3,y_2$ are in A.P.
• D. $y_1,y_3,y_2$ are in G.P.

Answer 1

Answer: (B), (C)

The correct options are
B $x_1,x_3,x_2$ are in G.P.
C $y_1,y_3,y_2$ are in A.P.

Let tangent $x-yt+a{t}^2=0$ intersect at $(x_3,y_3)$
Then $a{t}^2-y_{3}t+x_3=0$
now, $t_1+t_2=\frac{y_3}{a}$
$\Rightarrow 2a{t}_1+2a{t}_2=2y_3$
$\Rightarrow y_1+y_2=y_3$
Therefore, $y_1$, $y_3$, $y_2$ are in A.P

and $t_1{t}_2=\frac{x_3}{a}$
$\Rightarrow \left(a{t_1}^2\right)\left(a{t_2}^2\right)={x_3}^2$
$\Rightarrow x_1{x}_3={x_3}^2$
Therefore, $x_1$, $x_3$, $x_2$ are in G.P