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From a point P on the parabola \mathrm{y^2=4 d(x-a)}, tangents PQ and PR are drawn to the ellipse \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.} The locus of the mid-point of the chord QR is \mathrm{y^2=k d\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\left[x-a\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\right],} where \mathrm{k=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let (h, k) be a point on the given parabola \mathrm{y^2=4 d(x-a)}. The chord of contact of (h, k) with respect to the ellipse \mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1} is

                                 \mathrm{ \frac{\mathrm{xh}}{\mathrm{a}^2}+\frac{\mathrm{yk}}{\mathrm{b}^2}=1 }                                           .....(1)
The equation of the chord of the ellipse with mid point \mathrm{\left(\mathrm{x}_1, \mathrm{y}_{1)}\right.} is                  

   \mathrm{ \frac{x_1}{a^2}+\frac{y y_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} }                                                              .....(2)

(1) and (2) represent the same line. Hence                                                   

\mathrm{ \frac{\mathrm{h}}{\mathrm{x}_1}=\frac{\mathrm{k}}{\mathrm{y}_1}=\frac{1}{\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}} \Rightarrow \mathrm{h}=\frac{\mathrm{x}_1}{\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}}, \mathrm{k}=\frac{\mathrm{y}_1}{\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}} }
(h, k) lis in the parabola. Hence
\mathrm{ \frac{y_1^2}{\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)^2}=4 d\left(\frac{x_1}{\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}}-a\right) }
                                                         \mathrm{ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2\left[\frac{x}{\frac{x^2}{a^2}+\frac{y^2}{b^2}}-a\right] }

Hence the locus of (h, k) is \mathrm{ y^2=4 d}

\mathrm{ \Rightarrow y^2=4 d\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\left[x-a\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\right] \text {. } }

Posted by

Ritika Jonwal

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