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Consider the parabola \mathrm{y^2=8 x}; if the normal at a point \mathrm{P} on the parabola meets it again at a point \mathrm{Q}, then the least distance of \mathrm{Q}  from the tangent at the vertex of the parabola is

Option: 1

16


Option: 2

8


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

\mathrm{P \equiv\left(a t_1^2, 2 a t_1\right), Q \equiv\left(a t_2^2, 2 a t_2\right) \text {, then } t_2=-t_1-\frac{2}{t_1} }
Here  \mathrm{4 a=8 \quad \therefore a=2 }
\mathrm{z=a t_2^2=2\left(t_1^2+\frac{4}{t_1^2}+4\right)=2\left[\left(t_1-\frac{2}{t_1}\right)^2+8\right]\dots(i)}
Required distance,

\mathrm{ \Rightarrow z \geq 2 \text { (8) }}

\therefore  least value of  \mathrm{z=16}

Posted by

Ajit Kumar Dubey

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