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  • Find the locus of the point of intersection of the tangents to 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%20

Find the locus of the point of intersection of the tangents to the ellipse\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, if the difference of the eccentric angles of their points of contact is 2 a.
 

Option: 1

\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\sec ^2 \alpha}


Option: 2

\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos ^2 x}-


Option: 3

\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=\sec ^2 \alpha}-


Option: 4

\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=\cos ^2 \alpha}


Answers (1)

best_answer

Let the point of intersection be (h, k). Let \mathrm{\theta_1} and  \mathrm{\theta_2}be the eccentric angles such that

\mathrm{\theta_1-\theta_2=2 \alpha}

Points of contacts are  \mathrm{\left(a \cos \theta_1, b \sin \theta_1\right) and \left(a \cos \theta_2, b \sin \theta_2\right)} and the equations of tangents at these points are

\mathrm{\frac{x \cos \theta_1}{a}+\frac{y \sin \theta_1}{b}=1 \text { and } \frac{x \cos \theta_2}{a}+\frac{y \sin \theta_2}{b}=1}

Since these tangents pass through (h, k)


\mathrm{ \frac{h \cos \theta_1}{a}+\frac{k \sin \theta_1}{b}=1 \text { and } \frac{h \cos \theta_2}{a}+\frac{k \sin \theta_2}{b}=1} 

On solving these equations, we get
\mathrm{h=\frac{a\left(\sin \theta_1-\sin \theta_2\right)}{\sin \left(\theta_1-\theta_2\right)}=a \sec \alpha \cos \frac{\theta_1+\theta_2}{2}}

\begin{aligned} & \text { and }\mathrm{k=b \sec \alpha \sin \frac{\theta_1+\theta_2}{2} \Rightarrow \frac{h^2}{a^2} \cos ^2 \alpha+\frac{k^2}{b^2} \cos ^2 \alpha=1} \\ & \text { Required locus is } \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=\sec ^2 \alpha} \end{aligned}

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