Question

If the distance between two parallel tangents drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{49}=1$ is $2$, then their slope is equal to

• A. $\pm 5/2$
• B. $\pm 4/5$
• C. $\pm 7/2$
• D. $noneofthese$

Answer 1

The correct option is A $\pm 5/2$

$\frac{x^2}{9}-\frac{y^2}{49}=1$

Tangent $1\Rightarrow y=mx+\sqrt{a^2{m}^2-b^2}$

Tangent $2\Rightarrow y=mx-\sqrt{a^2{m}^2-b^2}$

Distance between two parallel lines = $\frac{|c_1-c_2|}{\sqrt{1+m^2}}$

$2=\frac{2\sqrt{a^2{m}^2-b^2}}{\sqrt{1+m^2}}$

$1+m^2=a^2{m}^2-b^2$

$1+b^2=(a^2-1)m^2$

$m=\sqrt{\frac{b^2+1}{a^2-1}}=\sqrt{\frac{49+1}{9-1}}=\sqrt{\frac{50}{8}}=\sqrt{\frac{25}{4}}=\pm \frac{5}{2}$