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  • A line intersects the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac%7

A line intersects the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at P and Q and the parabola \mathrm{y^2=4 d(x+a)} at R and S. The line segment PQ subtends a right angle at the centre of the ellipse. The locus of the point of intersection of the tangents to the parabola at R and S is \mathrm{y^2+4 d^2=k d^2(x+2 a)^2 \left(\frac{1}{a^2}+\frac{1}{b^2}\right)}, where \mathrm{k=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let (h, k) be the point of intersection of the tangents at R and S. The equation of the chord of contact of (h, k) w.r.t. the parabola \mathrm{y^2=4 d(x+a)} is \mathrm{k y=2 d(x+h)+4} ad i.e. ky \mathrm{-2 d x=4 \, a d+2 d h.}
This line subtends a right angle at the centre of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}. We homogenise the equation of the ellipse with the help of the equation of the chord. We get

\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\left(\frac{k y-2 d x}{4 a d+2 d h}\right)^2 }Since these lines are at right angle, we get.


.\mathrm{ \frac{1}{a^2}+\frac{1}{b^2}-\frac{k^2}{(4 a d+2 d h)^2}-\frac{4 d^2}{(4 a d+2 d h)^2}=0 }

Hence the locus of (h, k) is \mathrm{y^2+4 d^2=4 d^2(x+2 a)^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right).}

Posted by

Kuldeep Maurya

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