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  • Tangents are drawn at extremities of focal chord of a parabola <img alt="\mathrm{y^2=4 a x}" sr

Tangents are drawn at extremities of focal chord \mathrm{P Q} of a parabola \mathrm{y^2=4 a x} to intersect at \mathrm{R.} If \mathrm{G} is the centroid of \mathrm{\triangle P Q R} and \mathrm{O} is circumcentre, then \mathrm{O G} is parallel to the axis of parabola. Find the locus of mid-point of OG.

Option: 1

\mathrm{5 y^2+12 a x-8 a^2=0}


Option: 2

\mathrm{5 y^2-12 a x+8 a^2=0}


Option: 3

\mathrm{5 y^2+12 a x-4 a^{2}=0}


Option: 4

\mathrm{10 y^2-9 a x+4 a^2=0}


Answers (1)

best_answer

\mathrm{\text { Let } P(t) \, \, \& \, \, Q\left(-\frac{1}{t}\right)}

\mathrm{R=\left(-a, a\left(t-\frac{1}{t}\right)\right)}

\mathrm{G \equiv\left(\frac{a\left(t^2+\frac{1}{t^2}-1\right)}{3}, \frac{3 a\left(t-\frac{1}{t}\right)}{3}\right)}

Since tangents at extremities of a focal chord intersect at right angles so \mathrm{\triangle P Q R} is right angled at R. Hence circumcentre is mid point of \mathrm{P Q}.

\mathrm{\Rightarrow O \equiv\left(\frac{a\left(t^2+\frac{1}{t^2}\right)}{2}, \frac{2 a\left(t-\frac{1}{t}\right)}{2}\right)}

Since O and G have same ordinate so OG is parallel to axis.

\mathrm{ \Rightarrow A \equiv\left(\frac{\frac{a}{3}\left(t^2+\frac{1}{t^2}-1\right)+\frac{a}{2}\left(t^2+\frac{1}{t^2}\right)}{2}, a\left(t-\frac{1}{t}\right)\right) }

\mathrm{x=a{\left[\frac{5}{12}\left(t^2+\frac{1}{t^2}\right)-\frac{1}{6}\right]}, y=a{\left[t-\frac{1}{t}\right]}}

\mathrm{x=a{\left[\frac{5}{12}\left\{\left(t-\frac{1}{t}\right)2+2\right\}-\frac{1}{6}\right], t-\frac{1}{t}=\frac{y}{a}}}

on solving we get the locus as \mathrm{5 y^2-12 a x+8 a^2=0.}

Posted by

jitender.kumar

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