Question

If two tangents can be drawn to the different branches of hyperbola $\frac{x^2}{1}-\frac{y^2}{4}=1$ from $(a,a^2)$, then find the value of $a$

Answer 1

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

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

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

when $c=\pm \sqrt{A^2{m}^2-B^2}$

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

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

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

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

Squaring both sides, we get

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

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

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

For two real distinct values of $m,$ we get

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

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

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

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

$\Rightarrow {(a^2-2)}^2>0$

which is always greater than $0,$ except when $a^2-2=0$

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

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