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 : $\frac{x^2}{a^2}-\frac{y^2}{16}=1$
Given tangent
$2x-y+1=0$
using tangency condition
$m=2,1^2=a^2(2{)}^2-16\Rightarrow a=\frac{\sqrt{17}}{2}\cong 2.1$

From the options, using the pythagoras therom for the sides of a right angle triangle

$\text{A)}16\ne 1+\frac{17}{4}$

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

$\text{C)}64\ne 1+17$

$\text{D)}17=16+1$

Hence sides given in options $\text{(A), (B), (C)}$ will not form a right angle triangle.