Question

If two tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are drawn such that the product of their gradients is $c^2$, then they intersect at the curve

• A. $y^2+b^2=c^2(x^2-a^2)$
• B. $a{x}^2+b{y}^2=c^2$
• C. $y^2+b^2=c^2(x^2+a^2)$
• D. $y^2-b^2=c^2(x^2-a^2)$

Answer 1

The correct option is A $y^2+b^2=c^2(x^2-a^2)$

Let the tangents meets at the point $(h,k)$, then equation of tangents drawn from $(h,k)$ is given by

$S{S}_1=T^2$

$\Rightarrow (\frac{x^2}{a^2}-\frac{y^2}{b^2}-1)(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1)={(\frac{xh}{a^2}-\frac{yk}{b^2}-1)}^2$

$\Rightarrow x^2(\frac{h^2}{a^4}-\frac{k^2}{a^2{b}^2}-\frac{1}{a^2}-\frac{h^2}{a^4})-y^2(\frac{h^2}{a^2{b}^2}-\frac{k^2}{b^4}-\frac{1}{b^2}+\frac{k^2}{b^4})+...=0$

But $m_1{m}_2=\left(\frac{coefficientsof{x}^2}{coefficientsof{y}^2}\right)$

$\Rightarrow m_1{m}_2=\frac{\frac{k^2}{a^2{b}^2}+\frac{1}{a^2}}{\frac{h^2}{a^2{b}^2}-\frac{1}{b^2}}=c^2\Rightarrow \frac{(k^2+b^2)}{(h^2-a^2)}=c^2$

$\therefore y^2+b^2=c^2(x^2-a^2)$