Question

If the chord of contact of tangents from a point P to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends a right angle at the centre, then what is the locus of P?

• A. $x^2+y^2=0$
• B. $x^2+y^2=a^2$
• C. $x^2+y^2=a^2+b^2$
  
• D. $x^2+y^2=a^2-b^2$

Answer 1

Answer: (D)

$x^2+y^2=a^2-b^2$

Consider the given hyperbola.

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

We know that the line $y=mx+\sqrt{a^2{m}^2-b^2}$ is a tangent to the hyperbola.

Since, the chord of contact of tangents from point $P$.

Let the point $P(h,k)$.

Therefore,

$k=mh+\sqrt{a^2{m}^2+b^2}$

$k-mh=\sqrt{a^2{m}^2+b^2}$

On squaring both sides, we get

${(k-mh)}^2={\left(\sqrt{a^2{m}^2+b^2}\right)}^2$

$k^2+m^2{h}^2-2mkh=a^2{m}^2+b^2$

$m^2(h^2-a^2)-2mkh+(k^2+b^2)=0$

$m_1\cdot m_2=\frac{k^2+b^2}{h^2-a^2}$ ……. (2)

Since, a point P to the hyperbola subtends a right angle at the centre.

Therefore,

$\frac{k^2+b^2}{h^2-a^2}=-1$

$k^2+b^2=a^2-h^2$

$h^2+k^2=a^2-b^2$ …… (3)

On replacing $h,k$ by $x,y$, we get

$x^2+y^2=a^2-b^2$

Hence, the locus of $P$ is a circle.