Question

Locus of the pole of a chord of the hyperbola which subtends a right angles at the vertex is:

• A. $x=\frac{a^2+b^2}{a^2-b^2}$
• B. $x=\frac{a^2-b^2}{a^2+b^2}$
• C. $x=\frac{a^2+b^2}{a+b}$
• D. None of these

Answer 1

The correct option is D None of these

Let the equation of hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ......(i)

Vertex of the parabola is $(a,0)$

Shifting the origin to $(a,0)$, then equation of hyperbola becomes

$\frac{{(x+a)}^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=-\frac{2x}{a}\Rightarrow b^2{x}^2-a^2{y}^2=-2a{b}^{2}x$ ......(ii)

Let the pole of hyperbola be $(h,k)$

Equation of polar w.r.t to (i) $b^{2}hx-a^{2}ky=a^2{b}^2$

Shifting the origin to $(a,0)$

$\Rightarrow b^{2}h(x+a)-a^{2}ky=a^2{b}^2\Rightarrow b^{2}hx-a^{2}ky=a^2{b}^2-a{b}^{2}h\Rightarrow \frac{b^{2}hx-a^{2}ky}{a^2{b}^2-a{b}^{2}h}=1$ .........(iii)

Making the equation of hyperbola in (ii) homogeneous using equation of polar in (iii), we get

$b^2{x}^2-a^2{y}^2=-2a{b}^{2}x\left\{\frac{b^{2}hx-a^{2}ky}{a^2{b}^2-a{b}^{2}h}\right\}{b}^2{x}^2-a^2{y}^2+2x\left\{\frac{b^{2}hx-a^{2}ky}{a-h}\right\}=0$

Now the chord subtend right angle at the origin i.e. at $(a,0)$.

Therefore, coefficient of $x^2$ $+$ coefficient of $y^2=0$

$b^2-a^2+\frac{2b^{2}h}{a-h}=0-h(b^2-a^2)+a(b^2-a^2)+2b^{2}h=0\Rightarrow h=\frac{a(a^2-b^2)}{a^2+b^2}$

Generalizing the equation, we get

$\Rightarrow x=\frac{a(a^2-b^2)}{a^2+b^2}$

So, option D is correct.