Question

The value of m for which y = mx + 6 is a tangent to the hyperbola $\frac{x^2}{100}-\frac{y^2}{49}=1$ is .

• A. $\frac{\sqrt{17}}{10}$
• B. $\frac{\sqrt{17}}{20}$
• C. $\sqrt{\frac{17}{10}}$
• D. $\sqrt{\frac{17}{20}}$

Answer 1

The correct option is D $\sqrt{\frac{17}{20}}$

If y = mx + c touches $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,
then,
$c^2=a^2{m}^2-b^2\mathrm{Here},c=6,a^2=100\mathrm{and}{b}^2=49\therefore 36=100m^2-49\Rightarrow 100m^2=85\Rightarrow m=\pm \sqrt{\frac{17}{20}}$