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  • Prove that the product of the lengths of the perpendiculars drawn from the points (sqrt a^2-b^2,0) and (- sqrt a ^2-b^2,0) to the line x/a cos theta + y/b sin theta =1 is b^2.

Q: 23     Prove that the product of the lengths of the perpendiculars drawn from the points \small (\sqrt{a^2-b^2},0) and  \small (-\sqrt{a^2-b^2},0)  to the line  \small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1   is \small b^2.

Answers (1)

best_answer

Given equation id line is
\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1
We can rewrite it as
xb\cos \theta +ya\sin \theta =ab
Now, the distance of the line xb\cos \theta +ya\sin \theta =ab  from the point  \small (\sqrt{a^2-b^2},0)  is given by 
d_1=\left | \frac{b\cos\theta.\sqrt{a^2-b^2}+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
Similarly,
The distance of the line xb\cos \theta +ya\sin \theta =ab  from the point  \small (-\sqrt{a^2-b^2},0)  is given by 
d_2=\left | \frac{b\cos\theta.(-\sqrt{a^2-b^2})+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
d_1.d_2 = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |.\times\left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
           =\left | \frac{-((b\cos\theta.\sqrt{a^2-b^2})^2-(ab)^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
           =\left | \frac{-b^2\cos^2\theta.(a^2-b^2)+a^2b^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
           =\left | \frac{-a^2b^2\cos^2\theta+b^4\cos^2\theta+a^2b^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
           =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
           =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2(\sin^2\theta+\cos^2\theta))}{b^2\cos^2\theta+a^2\sin^2\theta} \right | \ \ \ \ (\because \sin^2a+\cos^2a=1)
           =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2\sin^2\theta-a^2\cos^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
           =\left | \frac{+b^2(b^2\cos^2\theta+a^2\sin^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
           =b^2
Hence  proved

Posted by

Gautam harsolia

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