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  • If the polar of a point w.r.t. <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;%5C

If the polar of a point w.r.t. \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} touches the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}, then the locus of the point is

Option: 1

Given hyperbola


Option: 2

Ellipse


Option: 3

Circle


Option: 4

None of these


Answers (1)

best_answer

Let \mathrm{(x_{1},y_{1})} be the given point.
Its polar w.r.t. \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \text { is } \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1 \quad \text { i.e. } \quad y=\frac{b^2}{y_1}\left(1-\frac{x x_1}{a^2}\right)=-\frac{b^2 x_1}{a^2 y_1} x+\frac{b^2}{y_1}}
This touches \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad \text { if } \quad\left(\frac{b^2}{y_1}\right)^2=a^2 \cdot\left(\frac{b^2 x_1}{a^2 y_1}\right)-b^2 \quad \Rightarrow \quad \frac{b^4}{y_1^2}=\frac{a^2 b^4 x_1^2}{a^4 y_1^2}-b^2 \Rightarrow}
\mathrm{\frac{b^2}{y_1^2}=\frac{b^2 x_1^2}{a^2 y_1^2}-1 \Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1}
\mathrm{\therefore } Locus of \mathrm{\left(x_1, y_1\right) \text { is } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}. Which is the same hyperbola.

Posted by

Ajit Kumar Dubey

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