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If P, Q, R are three points on a parabola \mathrm{y^2=4 a x} whose ordinates are in geometrical progression, then the trangents at P and R meet on

Option: 1

the line through Q parallel to x-axis


Option: 2

the line through Q parallel to y-axis 

 


Option: 3

the line joining Q to the vertex


Option: 4

the line joining Q to the focus 

 


Answers (1)

best_answer

Let the coordinates of P, Q, R be \mathrm{\left(a t_i^2, 2 a t_i\right) i=1,2,3}  having ordinates in G.P. So that \mathrm{t_1, t_2, t_3} are also in \mathrm{\text { G.P. i.e. } \mathrm{t}_1 \mathrm{t}_3=t_2^2}. Equations of the tangents at P and R are

\mathrm{t_1 y=x+a t_1^2 \text { and } t_3 y=x+a t_3^2 \text {, which intersect at the point } \frac{x+a t_1^2}{t_1}=\frac{x+a t_3^2}{t_3} \Rightarrow x a t_1 t_3=a t_2^2}

which is a line through Q parallel to Y-axis.

Posted by

Rakesh

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