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

Question

The equation of tangent for hyperbola is

$$y=mx\pm \sqrt {a^2m^2-b^2}$$

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

There will be two distinct values of $m$

$$4a^6-4(a^2-1)(a^4+4)>0$$ $$a^6-a^6-4a^2+a^4+4>0$$ $$a^4-4a^2+4>0$$ $$(a^2-2)^2>0$$ $$a^2>2$$ Or $$a^2<2$$

But the given answer for it is $a\in (-\infty, -2)\cup (2,\infty)$

Where have I gone wrong?

Answer 1

Your solution is correct

Let me verify by a separate method

We know two distinct tangents can be drawn for any point $P$ if it lies outside the hyperbola

So, we need $$\dfrac{a^2}1-\dfrac{a^4}4-1<0$$

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

$$\iff(a^2-2)^2>0$$ which is true if $a^2-2\ne0$ and real

Answer 2

Your reasoning works for $a^2\neq1$.

But for $a^2=1$ we obtain $(1,1)$ and $(-1,1)$, and they are valid!

Answer 3

With the given solution, I believe that the problem statement should read: “If two tangents can be drawn to different branches of a hyperbola...” This requires the $x$-coordinates of the tangency points to have opposite signs, and this indeed occurs for $a\in(-\infty,-2)\cup(2,\infty)$.

If you’re reproducing the problem statement correctly here, then there must be a typo in the original.