Question

If $2x-y+1=0$ is a tangent to the hyperbola$\frac{x^2}{a^2}-\frac{y^2}{16}=1$, then which of the following CANNOT be sides of a right triangle?

• A. $a,4,1$
• B. $2a,4,1$
• C. $a,4,2$
• D. $2a,8,1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

$a,4,1$

C

$a,4,2$

D

$2a,8,1$

Given the equation of tangent is $2x-y+1=0$
Hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{16}=1$
The point on the hyperbola be

$(a\sec \theta ,4\tan \theta )$
Tangent to the hyperbola in parametric form be $\frac{x\sec \theta }{a}-\frac{y\tan \theta }{4}=1$
On comparing, we get

$\sec \theta =2a$ and $\tan \theta =4\Rightarrow 4a^2-16=1$ [using trignometric identity]
$\therefore a=\frac{\sqrt{17}}{2}$