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  • Normals drawn 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%7Bx%5E2%7D%7Ba%5E2%7D&plus

Normals drawn to the ellipse  \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}  at point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid point of segment AB is 

 

Option: 1

\mathrm{4 x^2 a^2+4 y^2 b^2=\left(a^2-b^2\right)^2}


Option: 2

\mathrm{4 x^2 b^2+4 y^2 a^2=\left(a^2-b^2\right)}


Option: 3

\mathrm{16 x^2 a^2+16 y^2 b^2=\left(a^2-b^2\right)^2}


Option: 4

\mathrm{\quad 16 x^2 b^2+16 y^2 a^2=\left(a^2-b^2\right)}


Answers (1)

best_answer

Let  \mathrm{ P=(a \cos \theta, b \sin \theta)}  Equation of normal is,  \mathrm{ \frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2}
\mathrm{ \Rightarrow \quad A \equiv\left(\frac{a^2-b^2}{a} \cos \theta, 0\right) \text { and } B \equiv\left(0, \frac{a^2-b^2}{a} \sin \theta\right)}
It (h, k) be the mid– point of segment AB then 2h \mathrm{ =\frac{a^2-b^2}{a} \cos \theta, 2 k=\frac{a^2-b^2}{a} \sin \theta}

\mathrm{ \Rightarrow \quad 4 h^2 a^2+4 k^2 b^2=\left(a^2-b^2\right)^2}.

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jitender.kumar

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