Question

If $l\sin \theta -m\cos \theta =\sqrt{l^2+m^2}$ and $\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}=\frac{1}{l^2+m^2}$ then which of the following options is correct?

• A. $\frac{l^2}{b^2}+\frac{m^2}{a^2}=1$
• B. $\frac{l^2}{a^2}+\frac{m^2}{b^2}=1$
• C. $\frac{l^2}{b^2}-\frac{m^2}{a^2}=1$
• D. $\frac{l^2}{a^2}-\frac{m^2}{b^2}=1$

Answer 1

Answer: (A)

$l\sin \theta -m\cos \theta =\sqrt{l^2+m^2}$
Divide both sides by $\sqrt{l^2+m^2}$
$\left(\frac{1}{\sqrt{l^2+m^2}}\right)\sin \theta +\left(\frac{1-m}{\sqrt{l^2+m^2}}\right)\cos \theta =1$
$\Rightarrow {\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow \sin \theta \cdot \sin \theta +\cos \theta \cdot \cos \theta =1$
$\Rightarrow \left\{\because \sin \theta =\frac{l}{\sqrt{l^2+n^2}}\text{ and }\cos \theta =\frac{-m}{\sqrt{l^2+m^2}}\right\}$
$\Rightarrow \frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}=\frac{1}{l^2+m^2}$
Putting the value of $\sin \theta$ and $\cos \theta$, we get
$\frac{m^2}{\left(l^2+m^2\right)a^2}+\frac{l^2}{\left(l^2+m^2\right)b^2}=\frac{1}{l^2+a^2}$
Cancel the same terms on both sides.
$\frac{1}{l^2+m^2}\left(\frac{m^2}{a^2}+\frac{l^2}{b^2}\right)=\frac{1}{l^2+a^2}$
$\therefore \frac{m^2}{a^2}+\frac{l^2}{b^2}=1$