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If the polar with respect to \mathrm{y^2=4 a x} touches the ellipse  \mathrm{\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1} the locus of its pole is

Option: 1

\mathrm{\frac{x^2}{\alpha^2}-\frac{y^2}{\left(4 a^2 \alpha^2 / \beta^2\right)}=1}


Option: 2

\mathrm{\frac{x^2}{\alpha^2}+\frac{\beta^2 y^2}{4 a^2}=1}


Option: 3

\mathrm{\alpha^2 x^2+\beta^2 y^2=1}


Option: 4

None of these


Answers (1)

best_answer

Let\mathrm{P(h, k)} be the pole. Then the equation of the polar is \mathrm{k y=2 a(x+h) \text { or } y=\frac{2 a}{k} x+\frac{2 a h}{k}}
This touches \mathrm{\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1 \text {, So }\left(\frac{2 a h}{k}\right)^2=\alpha^2\left(\frac{2 a}{k}\right)^2+\beta^2} \mathrm{\text { (using } \left.c^2=a^2 m^2+b^2\right)}
\mathrm{\Rightarrow 4 a^2 h^2=4 a^2 \alpha^2+k^2 \beta^2 \text {. So, locus of }(h, k) \text { is } 4 a^2 x^2=4 a^2 \alpha^2+\beta^2 y^2 \text { or }}\mathrm{\frac{x^2}{\alpha^2}-\frac{y^2}{\left(\frac{4 a^2 \alpha^2}{\beta^2}\right)}=1}

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HARSH KANKARIA

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