Question

Evaluate y^ 2 b^ 2 - x^ 2 a^ 2 . where x = a tan 0 and y = b sec theta
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Answer 1

Answer:

The answer is $-a^2b^2=-(ab)^2$

Step-by-step explanation:

Given

$x=a\tan\theta\Rightarrow \tan\theta=\frac{x}{a}$ and $y=b\sec\theta\Rightarrow \sec\theta=\frac{y}{b}$

Now we know that $\sec^2\theta-\tan^2\theta=1$

                              $\Rightarrow (\frac{x}{a})^2-(\frac{y}{b})^2=1$

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