Question

If $\theta$ is eliminated from the equations $x = a \,cos( \theta - \alpha)$ and $y = b\,cos(\theta - \beta)$, then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} cos(\alpha - \beta)$ is equal to

• A. $sec^2 (\alpha - \beta)$
• B. $cosec^2(\alpha - \beta)$
• C. $cos^2(-\beta)$
• D. $sin^2(\alpha - \beta)$

Answer 1

Answer: (D)

$(\alpha - \beta) = (\theta - \beta) - (\theta - \alpha)$
$\Rightarrow cos(\alpha - \beta) = cos(\theta - \beta) cos(\theta - \alpha)$
$+ sin(\theta - \beta) sin(\theta - \alpha)$
$\Rightarrow \left[\frac{xy}{ab} - cos\left(\alpha - \beta\right)\right]^{2} =\left(1 - \frac{x^{2}}{a^{2}}\right)\left(1 - \frac{y^{2}}{b^{2}}\right)$
or $\frac{x^2y^2}{a^2b^2} + cos^2 (\alpha - \beta) - \frac{2xy}{ab} cos(\alpha -\beta)$
$ = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2y^2}{a^2b^2}$
or $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} cos(\alpha - \beta) = sin^2(\alpha - \beta)$