Question

If $m_{1}$and $m_{2}$ are the slopes of the tangents to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$ which pass through the point (6, 2) then

• A. $m_{1} + m_{2} = \frac{24}{11}$
• B. $m_{1}m_{2} = \frac{20}{11}$
• C. $m_{1} + m_{2} = \frac{48}{11}$
• D. $m_{1}m_{2} = \frac{11}{20}$

Answer 1

Answer: (B)

$m_{1}m_{2} = \frac{20}{11}$

Answer 2

⇒$\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$

Equation of tangent in terms of slope

$y = mx \pm \sqrt{\left( 25m^{2} - 16 \right)}$

or $(y - mx)^{2} = 25m^{2} - 16$it is passing through (6, 2) then

$(2 - 6m)^{2} = 25m^{2} - 16$⇒$4 + 36m^{2} - 24m = 25m^{2} - 16$⇒$11m^{2} - 24m + 20 = 0$$m_{1}m_{2} = \frac{20}{11}$