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 angled triangle?

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a$, 4, 1
B $a$, 4, 2
C $2a$, 8, 1

Hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{16}=1$
Tangent : $2x-y+1=0$
Tangent to a hyperbola,
$y=mx\pm \sqrt{a^2{m}^2-b^2}=mx\pm \sqrt{a^2{m}^2-16}$
Comparing them : $m=2$ and
$\sqrt{a^2{m}^2-16}=\pm 1$
Squaring both sides,
$\Rightarrow 4a^2-16=1\Rightarrow a=\frac{\sqrt{17}}{2}\approx 2.1$
From the options,
A) $16\ne 1+\frac{17}{4}$

B) $16\ne 4+\frac{17}{4}$

C) $64\ne 1+17$

D) $17=16+1$

Hence sides given in option A, B, C will not form a right angle triangle.