Question

Two tangents to the hyperbola $\frac{x^2}{100}-\frac{y^2}{81}=1$ having slopes $m_1$ and $m_2$ cuts the coordinate axes at four concyclic points. If $m_1$ and $m_2$ satisfy the equation $2{\alpha }^2-5\alpha +k=0,$ then the value of $k$ is

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

We have,
$\frac{x^2}{100}-\frac{y^2}{81}=1$
Equation of tangent having slope $m_1$ is
$L_1:y=m_{1}x+\sqrt{100m_1^2-81}$
Equation of tangent having slope $m_2$ is
$L_2:y=m_{2}x+\sqrt{100m_2^2-81}$

Coordinates of $A\equiv \left(\frac{-\sqrt{100m_1^2-81}}{m_1},0\right)$
Coordinates of $C\equiv \left(0,\sqrt{100m_1^2-81}\right)$
Coordinates of $B\equiv \left(\frac{-\sqrt{100m_2^2-81}}{m_2},0\right)$
Coordinates of $D\equiv \left(0,\sqrt{100m_2^2-81}\right)$
From the property of cyclic quadrilateral,
$OA\cdot OB=OC\cdot OD$
$\Rightarrow \left[\left(\frac{-\sqrt{100m_1^2-81}}{m_1}\right)\times \left(\frac{-\sqrt{100m_2^2-81}}{m_2}\right)\right]=\left[\sqrt{100m_1^2-81}\times \sqrt{100m_2^2-81}\right]$
$\Rightarrow m_1{m}_2=1$
Now, $m_1$ and $m_2$ are satisfying $2{\alpha }^2-5\alpha +k=0$ , therefore
$\frac{k}{2}=m_1{m}_2=1$
$\Rightarrow k=2$