Question

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

• A. Given hyperbola
• B. Ellipse
• C. Circle
• D. None of these

Answer 1

Answer: (A)

Given hyperbola

Answer 2

Let $(x_{1},y_{1})$ be the given point.

Its polar w.r.t. $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $\frac{xx_{1}}{a^{2}} + \frac{yy_{1}}{b^{2}} = 1$ i.e.,

$y = \frac{b^{2}}{y_{1}}\left( 1 - \frac{xx_{1}}{a^{2}} \right) = - \frac{b^{2}x_{1}}{a^{2}y_{1}}x + \frac{b^{2}}{y_{1}}$This touches $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

if $\left( \frac{b^{2}}{y_{1}} \right)^{2} = a^{2}.\left( \frac{b^{2}x_{1}}{a^{2}y_{1}} \right) - b^{2}$ ⇒ $\frac{b^{4}}{y_{1}^{2}} = \frac{a^{2}b^{4}x_{1}^{2}}{a^{4}y_{1}^{2}} - b^{2}$

⇒ $\frac{b^{2}}{y_{1}^{2}} = \frac{b^{2}x_{1}^{2}}{a^{2}y_{1}^{2}} - 1$ ⇒ $\frac{x_{1}^{2}}{a^{2}} - \frac{y_{1}^{2}}{b^{2}} = 1$

$\therefore$ Locus of $(x_{1},y_{1})$ is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.

Which is the same hyperbola