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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 $ab$.

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    The slope of $(xx_1/(a^2-b^2))-(yy_1/(a^3-b^3))-1=0$ is $x_1(a^3-b^3)/(y_1(a^2-b^2))$, not $(a^3-b^3)/(a^2-b^2)$. – 2017-02-22
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    Oh! Right. How could I have missed that! Is there any way we can now establish the inequality? – 2017-02-22

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Is there any way we can now establish the ineuqality?

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.

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    This is not the extension to my solution. It is a bit lengthy. I'd rather use $y=mx±(A^2m^2-b^2)^1/2$ for hyperbola $x^2/A^2-y^2/B^2$ and put $A^2m^2-b^2≥ 0$ – 2017-02-22
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    @Cotton: Yeah, that must be better. The answer contains the proof that the tangent line with slope $m$ is written so. – 2017-02-22