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

Answer 1

The correct option is A $2a,4,1$

Given: Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$ -------------(1)

Tangent : $2x-y+1=0$------2 or $2x+1=y$ ---------------(2)

As we know, general form of tangent is $y=mx+c$ where condition of tangency $\Rightarrow c^2=a^2{m}^2-b^2$ -------------(3)

From Equation of Hyperbola, $a=a,b=4$

And values of $c$ and $m$ from the given equation of tangent, $c=1,m=2$

$\Rightarrow$ In Equation (3), $\Rightarrow 1^2=a^2(2{)}^2-(4{)}^2$

$\Rightarrow 1^2=4a^2-16$

$\Rightarrow 4a^2=17\Rightarrow a=\frac{\pm \sqrt{1}7}{2}$

Now, checking the options with Pythagoras Theorem as they are sides of right angled triangle,

A) $2a,4,1\Rightarrow \sqrt{1}7,4,1\Rightarrow \sqrt{1}7>\sqrt{1}6\Rightarrow \sqrt{1}7>4$

So, $\sqrt{1}7$ is the bigger side.

By Pythagoras theorem, $(1{)}^2+(4{)}^2=(\sqrt{1}7{)}^2$

$1+16=17$ It is a correct set.

B) $2a,8,1$

$8>\sqrt{1}7\Rightarrow 8$ is bigger side.

$\Rightarrow 64=(\sqrt{1}7{)}^2+(1{)}^2$

$\Rightarrow 64\ne 17+1$

So, they cannot be sides of the triangle.

C) $a,4,1$

$4>\frac{\sqrt{1}7}{2}\Rightarrow 4$is the biggest side.

$16=\frac{17}{4}+1$

$16\ne \frac{21}{4}$

They cannot be sides of a right angled triangle.

D) $9,4,2$

$4>\frac{\sqrt{1}7}{2}$

$4^2=(\frac{\sqrt{1}7}{2}{)}^2+(2{)}^2=\frac{17}{4}+4$

$\Rightarrow 16\ne \frac{33}{4}$

They cannot be sides of a right angled triangle.