Question

If two tangent can be drawn to the different breaches of hyperbola $\frac{x^2}{1}-\frac{y^2}{4}=1$ from the point $\left(a,a^2\right)$ then

• A. $a\in \left(-2,0\right)$
• B. $a\in \left(0,2\right)$
• C. $a\in \left(-\infty ,-2\right)$
• D. $a\in \left(2,\infty \right)$

Answer 1

The correct option is D $a\in \left(2,\infty \right)$

$\frac{x^2}{1}-\frac{y^2}{4}=1$

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

on comparing with $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$, we get $A=1$ & $B=2$

$y=mx+c$ is tangent to $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ where $c=\pm \sqrt{A^2{m}^2-B^2}$

$\Rightarrow y=mx\pm \sqrt{A^2{m}^2-B^2}$

As tangent passes through $(a,a^2)$

$\Rightarrow a^2=ma\pm \sqrt{m^2-4}$

$\Rightarrow a^2-am=\pm \sqrt{m^2-4}$

$\Rightarrow (a^2-am{)}^2=m^2-4$ (squaring both sides)

$\Rightarrow a^4+a^2{m}^2-2a^{3}m=m^2-4$

$\Rightarrow m^2(1-a^2)+2a^{3}m-(a^4+4)=0$

For two real distinct value of m, we get

$(2a^3{)}^2-4(1-a^2)\{-(a^4+4\left)\right\}>0$ $\{\because b^2-4ac>0\}$

$\Rightarrow a^6+(1-a^2)(a^4+4)>0$

$\Rightarrow a^6+a^4+4-a^6-4a^2>0$

$\Rightarrow a^4-4a^2+4>0$

$\Rightarrow (a^2-2{)}^2>0$ which is always greater than $0$, except when $a^2-2=0$

$\Rightarrow a^2-2\ne 0$

$\Rightarrow a=\pm \sqrt{2}$

$a\sim \pm 1.414$ only option (d) doesnot contain $\pm 1.414$

Hence, option (d) is correct.