Prove that the product of the lengths of the perpendiculars drawn from the points.

$$\sqrt{(a^2-b^2,0)}, \sqrt{-(a^2-b^2,0)},$$  to the line $$\dfrac{x}{a}cos θ +\dfrac{y}{b}sinθ=1$$ is b2

Asked by Abhisek | 1 year ago |  143

##### Solution :-

It is given that

$$\dfrac{x}{a}cos θ +\dfrac{y}{b}sinθ=1$$

We can write it as

bx cos θ + ay sin θ – ab = 0 ….. (1)

Here, the length of the perpendicular from point $$\sqrt{(a^2-b^2,0)}$$ to line (1) is,

$$p_1=\dfrac{|bcos θ(\sqrt{a^2-b^2}+asin θ(0)-ab)|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}$$

$$p_1=\dfrac{|bcos θ(\sqrt{a^2-b^2}-ab)|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}$$

Similarly, the length of the perpendicular from point$$\sqrt{-(a^2-b^2,0)},$$  to the line (2) is,

$$p_2=\dfrac{|bcos θ(-\sqrt{a^2-b^2}+asin θ(0)-ab)|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}$$

$$p_2=\dfrac{|bcos θ(\sqrt{a^2-b^2}+ab)|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}$$

Multiply equations (2) and (3),

From the formula,

$$\dfrac{|bcos θ(\sqrt{a^2-b^2}-(ab)^2)|}{b^2cos^2θ+a^2sin^2θ}$$

Square the numerator,

$$\dfrac{|b^2cos^2 θ(a^2-b^2)-a^2b^2|}{b^2cos^2θ+a^2sin^2θ}$$

Expand using formula

$$\dfrac{|a^2b^2cos^2 θ-b^4cos^2θ-a^2b^2|}{b^2cos^2θ+a^2sin^2θ}$$

Take out the common terms,

$$b^2\dfrac{|a^2cos^2 θ-b^4cos^2θ-a^2|}{b^2cos^2θ+a^2sin^2θ}$$

$$b^2\dfrac{|a^2cos^2 θ-b^2cos^2θ-a^2sin^2θ-a^2cos^2θ|}{b^2cos^2θ+a^2sin^2θ}$$

= Here,sin2θ+cos2θ=1 (trigonometric identity)

$$b^2\dfrac{|-b^2cos^2 θ+a^2sin^2θ|}{b^2cos^2θ+a^2sin^2θ}$$

$$b^2\dfrac{|b^2cos^2 θ+a^2sin^2θ|}{b^2cos^2θ+a^2sin^2θ}$$

Cancel common terms,
$$b^2$$
Hence, proved that the product of the lengths of the perpendiculars drawn from the points

Answered by Pragya Singh | 1 year ago

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