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The tangents to the parabola y^2=4a x at...

The tangents to the parabola `y^2=4a x` at the vertex `V` and any point `P` meet at `Q` . If `S` is the focus, then prove that `S P,S Q ,` and `S V` are in GP.

Text Solution

Verified by Experts

Let the parabola be `y^(2)=4ax`.
Q is the intersection of the line x=0 and the tangent at point
`P(at^(2),2at),ty=x+at^(2)`.
Solving these, we get Q = (0, at). Also, S=(a,0).
Now, focal length
`SP=a+at^(2)`
`SQ^(2)=a^(2)+a^(2)t^(2)=a^(2)(t^(2)+1)`
and SV = a
`:." "SQ^(2)=SPxxSV`
Therefore, SP,SQ and SV are in GP.
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