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  • An ordinate PN of an 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%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%

An ordinate PN of an ellipse \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1} meets the auxiliary circle in Q. The locus of the point of intersection of the normals at P and Q is the circle \mathrm{x^2+y^2=k^2.}

Option: 1

\mathrm{a-b}


Option: 2

\mathrm{2b-a}


Option: 3

\mathrm{a+b}


Option: 4

\mathrm{None \, \, of \, \, these}


Answers (1)

best_answer

Let a ordinate be \mathrm{y=b \sin \theta}

\mathrm{P \equiv(a \cos \theta, b \sin \theta) \, \, and \, \, Q \equiv(a \cos \theta, a \sin \theta)}

Equation of normal at P is ax \mathrm{\sec \theta-by \operatorname{cosec} \theta=a^2-b^2}          ....(1)

Equation of normal at Q is \mathrm{x \sin \theta-y \cos \theta=0}                               ....(2)

Let the point of intercession of (1) and (2) be \mathrm{(\alpha, \beta)}

\mathrm{\text { Then } \tan \theta=\frac{\beta}{\alpha} \Rightarrow \sec \theta=\frac{\sqrt{\alpha^2+\beta^2}}{\alpha}, \operatorname{cosec} \theta=\frac{\sqrt{\alpha^2+\beta^2}}{\beta}}

Putting these values in (1) we have

\mathrm{ \begin{aligned} & \text { a } \alpha \frac{\sqrt{\alpha^2+\beta^2}}{\alpha}-b \beta \frac{\sqrt{\alpha^2+\beta^2}}{\beta}=a^2-b^2 \\\\ & \Rightarrow \sqrt{\alpha^2+\beta^2}(a-b)=(a-b)(a+b) \Rightarrow a^2+\beta^2=(a+b)^2 \end{aligned} }

Hence required locus will be \mathrm{x^2+y^2=(a+b)^2.}

Posted by

Divya Prakash Singh

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