Given hyperbola: $$\left(\frac{x^2}{a^2-b^2}\right)-\left(\frac{y^2}{a^3-b^3}\right)=1$$
Amswer given is (1).
I only got the equality part of the given answer. I don't know how inequality will be formed in my solution. Did I miss something in my solution?
Please let me know the flaw in my procedure and not other possible solutions to this problem.
In my attempt, in the second and third step, it is $a+b$ in the denominator and not $ab$.
Let $y=mx+n$ be the tangent line.
Eliminating $y$ gives $$\frac{x^2}{a^2-b^2}-\frac{(mx+n)^2}{a^3-b^3}=1,$$ i.e. $$\left(\frac{1}{a^2-b^2}-\frac{m^2}{a^3-b^3}\right)x^2-\frac{2mn}{a^3-b^3}x-\frac{n^2}{a^3-b^3}-1=0$$
Since the discriminant has to be $0$, we get $$\left(\frac{-2mn}{a^3-b^3}\right)^2-4\left(\frac{1}{a^2-b^2}-\frac{m^2}{a^3-b^3}\right)\left(-\frac{n^2}{a^3-b^3}-1\right)=0,$$ i.e. $$\frac{m^2}{a^3-b^3}=\frac{1}{a^2-b^2}+\frac{n^2}{(a^2-b^2)(a^3-b^3)}$$ from which we get $$\frac{m^2}{a^3-b^3}\ge \frac{1}{a^2-b^2}$$ So, $(a)$ follows.