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  • The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angles differ by a constant <img alt="\mathrm{\alpha}" src="https://entrancecorner.oncodecogs.com/gif.

The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angles differ by a constant \mathrm{\alpha} is an ellipse \mathrm{b^2 x^2+a^2 y^2=k^2 \sec ^2\left(\frac{\alpha}{2}\right)}., where \mathrm{k=}

Option: 1

\mathrm{\sqrt{a b}}


Option: 2

\mathrm{a b}


Option: 3

\mathrm{\frac{a}{b}}


Option: 4

\mathrm{\frac{b}{a}}


Answers (1)

best_answer

Equation of tangent

\mathrm{ \frac{\mathrm{x}}{\mathrm{a}} \cos \varphi+\frac{\mathrm{y}}{\mathrm{b}} \sin \varphi=1 }                                ......(1)

\mathrm{ \frac{\mathrm{x}}{\mathrm{a}} \cos (\varphi+\alpha)+\frac{\mathrm{y}}{\mathrm{b}} \sin (\varphi+\alpha)=1 }        ......(2)
As these intersect at R(h, k)

\mathrm{\therefore \mathrm{bh} \cos \varphi+\mathrm{ak} \sin \varphi-\mathrm{ab}=0}

\mathrm{bh \, \, \cos (\varphi+\alpha)+a k \sin (\varphi+\alpha)-a b=0}
Solving together

\mathrm{ \begin{aligned} & \frac{\mathrm{bh}}{-\sin \phi+\sin (\phi+\alpha)}=\frac{\mathrm{ak}}{-\cos (\phi+\alpha)+\cos \phi}=\frac{\mathrm{ab}}{\sin \alpha} \\\\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\cos \left(\frac{\phi+\phi+\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \frac{\mathrm{k}}{\mathrm{b}}=\frac{\sin \left(\frac{\phi+\phi+\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \end{aligned} }
\therefore Squaring and adding

\mathrm{ \frac{\mathrm{h}^2}{\mathrm{a}^2}+\frac{\mathrm{k}^2}{\mathrm{~b}^2}=\sec ^2 \frac{\alpha}{2} }

Hence locus is

\mathrm{b^2 x^2+a^2 y^2=a^2 b^2 \sec ^2\left(\frac{\alpha^2}{2}\right) .}

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