Question

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

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

Answer 1

Answer: (C)

$(x^2+y^2{)}^2+4a^2(x^2-y^2)=4a^4$

Given equation of hyperbola is
$x^2-y^2=a^2$
or $\frac{x^2}{a^2}-\frac{y^2}{a^2}=1$
Let $y=mx\pm \sqrt{m^2{a}^2-a^2}$ be two tangents to the hyperbola.
Since, it passes through $(h,k).$
$\Rightarrow (k-mh{)}^2=m^2{a}^2-a^2$
$\Rightarrow m^2(h^2-a^2)-2khm+k^2+a^2=0$
$\Rightarrow m_1+m_2=\frac{2kh}{h^2-a^2}$and $m_1{m}_2=\frac{k^2+a^2}{h^2-a^2}$
Now, $tan{45}^o=\frac{m_1-m_2}{1+m_1{m}_2}$
$\Rightarrow 1=\frac{(m_1+m_2{)}^2-4m_1{m}_2}{(1+m_1{m}_2{)}^2}$
$\Rightarrow (1+\frac{k^2+a^2}{h^2-a^2}{)}^2=(\frac{2kh}{h^2-a^2}{)}^2-4\left(\frac{k^2+a^2}{h^2-a^2}\right)$
$\Rightarrow (h^2+k^2{)}^2=4h^2{k}^2-4(k^2+a^2\left)\right(h^2-a^2)$
$\Rightarrow (x^2+y^2{)}^2=4(a^2{y}^2-a^2{x}^2+a^4)$
$\Rightarrow (x^2+y^2{)}^2+4a^2(x^2-y^2)=4a^4$