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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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 $ \large\frac{x}{a}$$ \cos \theta + \large\frac{y}{b}$$ \sin \theta = 1$ is $b^2$

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Toolbox:
  • Length of the perpendicular from a point $(x_1, y_1)$ is $ d = \large\frac{|Ax_1+By_1+c|}{\sqrt{A^2+B^2}}$
The given equation is
$ \large\frac{x}{a}$$ \cos \theta + \large\frac{y}{b}$$ \sin \theta = 1$
This can be written as
$ bx \cos \theta + ay \sin \theta = ab$
or $ bx \cos \theta + ay \sin \theta - ab = 0$------(1)
$ \therefore $ Length of the perpendicular from the point $ ( \sqrt{a^2-b^2}$$, 0) $ is
$ p_1 = \large\frac{|b \cos \theta ( \sqrt{a^2-b^2}+ a \sin \theta (0) - ab|}{\sqrt{b^2 \cos^2 \theta + a^2 sin^2 \theta}}$
$ = \large\frac{|b \cos \theta \sqrt{a^2-b^2}-ab|}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}}$-------(2)
Length of the perpendicular from point $ ( - \sqrt{a^2-b^2}, 0)$ is
$ p_2 = \large\frac{| b \cos \theta ( -\sqrt{a^2-b^2})+ a \sin \theta (0)-ab |}{\sqrt{b^2 \cos^2\theta + a^2 \sin^2 \theta}}$
$ = \large\frac{b \cos \theta \sqrt{a^2-b^2}+ ab|}{\sqrt{b^2\cos^2 \theta+a^2\sin^2 \theta}}$-------(3)
Multiplying equation (2) and (3)
$p_1p_2 = \large\frac{b \cos \theta \sqrt{a^2-b^2}-ab||b \cos \theta \sqrt{a^2-b^2}+ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}\sqrt{b^2\cos^2 \theta+a^2\sin^2 \theta}}$
$ = \large\frac{|b^2\cos^2\theta (a^2-b^2)-a^2b^2|}{b^2\cos^2\theta+a^2\sin^2\theta}$
$ = \large\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}$
$ = \large\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}$
But $(\cos^2\theta - 1 = -\sin^2 \theta)$
$ = \large\frac{b^2|-a^2\sin^2\theta - b^2\cos^2 \theta|}{b^2 \cos^2 \theta + a^2\sin^2 \theta}$
$= \large\frac{b^2(a^2 \sin^2 \theta + b^2 \cos^2 \theta )}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$
$ = b^2$
Hence proved.
answered May 23, 2014 by thanvigandhi_1
 

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