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  • - The locus of the point of intersection of 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%5Cfrac%7B

- The locus of the point of intersection of tangents to the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at the points whose eccentric angles differ by \mathrm{\pi / 2} is

Option: 1

\mathrm{x^2+y^2=a^2}


Option: 2

\mathrm{x^2+y^2=b^2}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\text { Let } P(a \cos \theta, b \sin \theta) \text { and } Q(a \cos (\pi / 2+\theta), b \sin (\pi / 2+\theta))}

be two points on the ellipse. The equations of tangents to the ellipse at points P and Q are

\mathrm{\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1}              ...[i]

\mathrm{\text { and }-\frac{x}{a} \sin \theta+\frac{y}{b} \cos \theta=1}       ....[ii]

Let R(h, k) be the point of intersection of (i) and (ii). Then,

\mathrm{\begin{aligned} & \frac{h}{a} \cos \theta+\frac{k}{b} \sin \theta=1 \text { and }-\frac{h}{a} \sin \theta+\frac{k}{b} \cos \theta=1 \\ & \Rightarrow\left(\frac{h}{a} \cos \theta+\frac{k}{b} \sin \theta\right)^2+\left(-\frac{h}{a} \sin \theta+\frac{k}{b} \cos \theta\right)^2=2 \\ & \Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=2 \end{aligned}}

\mathrm{\text { Hence, the locus of }(h, k) \text { is } \frac{x^2}{a^2}+\frac{y^2}{b^2}=2}

Posted by

Suraj Bhandari

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