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From a point O on the circle \mathrm{x^2+y^2=25,}  tangents OP and OQ are drawn to the ellipse \mathrm{\frac{x^2}{4}+\frac{y^2}{1}=1.} The locus of the mid-point of the chord PQ describes the curve \mathrm{x^2+y^2=k\left[\frac{x^2}{4}+\frac{y^2}{1}\right]^2}, where \mathrm{ k=}

Option: 1

5


Option: 2

25


Option: 3

15


Option: 4

4


Answers (1)

best_answer

Let \mathrm{O \equiv(5 \cos \theta, 5 \sin \theta)}

Let R be middle point of PQ. Let \mathrm{R \equiv(h, k).}

Equation of the chord of contact PQ is \mathrm{T=0}

i.e. \mathrm{\frac{5 x \cos \theta}{4}+\frac{5 y \sin \theta}{1}=1}                             ......(1)

Equation of the chord PQ with middle point R(h, k) is

\mathrm{T=S_1\, \, i.e. \frac{x h}{4}+\frac{y k}{1}-1=\frac{h^2}{4}+\frac{k^2}{1}-1 \, \, or \, \, \frac{x h}{4}+\frac{y k}{1}=\frac{h^2}{4}+\frac{k^2}{1}}           ......(2)
(1) and (2) are the same equations.

\mathrm{ \begin{aligned} & \quad \frac{5 \cos \theta}{h}=\frac{5 \sin \theta}{k}=\frac{1}{\frac{h^2}{4}+\frac{k^2}{1}} \Rightarrow \frac{\left(h^2+k^2\right)}{\left(\frac{h^2}{4}+\frac{k^2}{1}\right)^2} \cdot \frac{1}{25}=1 \\ & \Rightarrow \\ & \text { Hence locus of }(h, k) \text { is }\left(x^2+y^2\right)=25\left[\frac{x^2}{4}+\frac{y^2}{1}\right]^2 . \end{aligned} }

Posted by

Kuldeep Maurya

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