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If $(-2,5)$ and $(3,7)$ are the points of intersection of the tangent and normal at a point on a parabola $\mathrm{y^{2}=4ax}$ with the axis of the parabola ,then the focal distance of that point is
Option: 1
$\frac{\sqrt29}{2}$

Option: 2
$\frac{5}{2}$

Option: 3
$\sqrt29$

Option: 4
$\frac{2}{5}$

Answers (1)

best_answer

For parabola $\mathrm{y^2=4 a x}$ , tangent and normal at point $\mathrm{P\left(a t^2, 2 a t\right)}$ meets x-axis at $\mathrm{T\left(-a t^2, 0\right)\: and \: N\left(2 a+a t^2, 0\right)}$ Thus, focus $\mathrm{S}$ is the mid-point of $\mathrm{TN}$

Also, $\mathrm{S P=S T=S N=a+a t^2}$

$\mathrm{ \therefore S P=T N / 2 }$

For given data, $\mathrm{ T N=\sqrt{29} }$

$\mathrm{ \therefore S P=\frac{\sqrt{29}}{2} }$

Hence option 1 is correct.

Posted by

Kuldeep Maurya

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