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
$L_{1}:y=m_{1}x+\sqrt{100 m_{1}^{2} - 81}$
Equation of tangent having slope $m_{2}$ is
$L_{2}:y=m_{2}x+\sqrt{100 m_{2}^{2} - 81}$

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