Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A circle is drawn considering any focal chord of the parabola <img alt="\mathrm{y^2=4 a x }"

A circle is drawn considering any focal chord \mathrm{P Q }of the parabola \mathrm{y^2=4 a x }as diameter which intersects the tangent at \mathrm{P^{\left(a t^2, 2 a t\right)} } in \mathrm{A }. Then the length of the side \mathrm{A Q } of \mathrm{\triangle A S Q } is (where S is the focus of the parabola)
 

Option: 1


\mathrm{\frac{a}{t^2}\left(1+t^2\right)^{3 / 2}}


Option: 2

\mathrm{a t^2\left(1+t^2\right)^{3 / 2}}


Option: 3

\mathrm{\frac{a t^2}{\left(1+t^2\right)^{3 / 2}}}


Option: 4

none of these


Answers (1)

 Required distance is perpendicular distance of Q from tangent at P. Equation of tangent at P is

\begin{aligned} &\mathrm{ x-t y+a t^2=0 } \quad \quad \quad \quad \left[\begin{array}{ll} &\mathrm{ Q:\left(\frac{a}{t^2},-\frac{2 a}{t}\right)} \end{array}\right]\\ &\text{ Required distance =}\mathrm{ \left|\frac{\frac{a}{t^2}-t \times\left(-\frac{2 a}{t}\right)+a t^2}{\sqrt{1+t^2}}\right|=\frac{a}{t^2} \frac{\left(1+t^2\right)^2}{\sqrt{\left(1+t^2\right)}}=\frac{a}{t^2}\left(1+t^2\right)^{3 / 2} }\\ \end{aligned}

Posted by

Kshitij

View full answer

Similar Questions

Latest Question