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  • The straight line joining any point on the parabola <img alt="\mathrm{\mathrm{y}^2=4 \mathrm{ax}}" src

The straight line joining any point \mathrm{P}on the parabola \mathrm{\mathrm{y}^2=4 \mathrm{ax}} to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equaiton of the locus of R is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\mathrm{T}: \mathrm{ty}=\mathrm{x}+\mathrm{at} \mathrm{t}^2}                            \mathrm{...(1)}

line perpendicular to (1) through (a, 0)

\mathrm{t x+y=t a }                                         \mathrm{...(2)}

\mathrm{\text { equation of } \mathrm{OP}: \mathrm{y}-\frac{2}{\mathrm{t}} \mathrm{x}=0}          \mathrm{...(3)}

from (2) & (3) eleminating t we get locus

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vinayak

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