Question

The locus of a point, from where pair of tangents to the rectangular hyperbola $x^2-y^2=a^2$ contain an angle of ${45}^{\circ }$, is :

• A. $(x^2+y^2{)}^2+4a^2(x^2-y^2)=4a^4$
• B. $(x^2+y^2{)}^2+4a^2(x^2-y^2)=a^4$
• C. $2(x^2+y^2)+4a^2(x^2-y^2)=4a^2$
• D. $(x^2+y^2)+a^2(x^2-y^2)=4a^2$

Answer 1

The correct option is A $(x^2+y^2{)}^2+4a^2(x^2-y^2)=4a^4$

Equation of tangent to given hyperbola is : $y=mx\pm \sqrt{a^2{m}^2-a^2}$
$\Rightarrow$ Let $P(h,k)$ be the locus
$\Rightarrow k-mh=\pm \sqrt{a^2{m}^2-a^2}$
$\Rightarrow k^2+m^2{h}^2-2kmh=a^2{m}^2-a^2$
$\Rightarrow m^2(h^2-a^2)-2hkm+k^2+a^2=0$
$\therefore m_1+m_2=\frac{2hk}{h^2-a^2}$ and
$m_1{m}_2=\frac{k^2+a^2}{h^2-a^2}$
Now,
$\tan {45}^{\circ }=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$\Rightarrow (m_1-m_2{)}^2=(1+m_1{m}_2{)}^2$
$\Rightarrow (m_1+m_2{)}^2-4m_1{m}_2=(1+m_1{m}_2{)}^2$
$\Rightarrow {\left(\frac{2hk}{h^2-a^2}\right)}^2-4\left(\frac{k^2+a^2}{h^2-a^2}\right)={\left(\frac{h^2+k^2}{h^2-a^2}\right)}^2$
$\Rightarrow (h^2+k^2{)}^2+4a^2(h^2-k^2)=4a^4$