Question

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

• A. $\alpha ϵ(-2,0)$
• B. $\alpha ϵ(2,0)$
• C. $\alpha ϵ(-\infty ,2)$
• D. $\alpha ϵ(,2-\infty )$

Answer 1

The correct option is D $\alpha ϵ(,2-\infty )$

We have,

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

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

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

We get,

$A=1,B=2$

Then, we know that equation of tangent in parabola,

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

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

Now,

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

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

Squaring both side and we get,

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

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

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

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

For two real distinct values of m, we get

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

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

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

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

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

Now, which is always greater than zero, except when

$a^2-2=0$

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

$\Rightarrow a\approx \pm 1.414$

Hence, $\alpha \in (2,-\infty )$ the answer.