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

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