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  • If the points of intersection of the ellipses <img alt="\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac%7Bx%5E2%7D%7Ba%

If the points of intersection of the ellipses \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1} and \mathrm{\frac{x^2}{p^2}+\frac{y^2}{q^2}=1} be the extremities of the conjugate diameters of first ellipse, then \mathrm{\frac{a^2}{p^2}+\frac{b^2}{q^2}=}

Option: 1

0


Option: 2

2


Option: 3

3


Option: 4

1


Answers (1)

best_answer

: Subtracting in order to find their points of intersection, we get

\mathrm{x^2\left(\frac{1}{a^2}-\frac{1}{p^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{q^2}\right)=0}

Above equation will represent a pair of conjugate diameters of first ellipse if

\mathrm{m_1 m_2=-\frac{b^2}{a^2}}

\mathrm{\text { But } m_1 m_2=\frac{A}{B}=-\frac{b^2}{a^2}}

\mathrm{\begin{aligned} & \therefore\left(\frac{1}{a^2}-\frac{1}{p^2}\right) \div\left(\frac{1}{b^2}-\frac{1}{q^2}\right)=-\frac{b^2}{a^2} \\ & \text { or } a^2\left(\frac{1}{a^2}-\frac{1}{p^2}\right)+b^2\left(\frac{1}{b^2}-\frac{1}{q^2}\right)=0 \\ & \text { or } \frac{a^2}{p^2}+\frac{b^2}{q^2}=2 \end{aligned}}

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Shailly goel

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