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Question Number 75484 by vishalbhardwaj last updated on 11/Dec/19

If θ is eleminated from the

equation x = a c o s (θ - α) and

y = b c o s (θ - β) then prove that

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{2 x y}{a b} c o s (α - β)

= {s i n}^{2} (α - β)

Answered by MJS last updated on 12/Dec/19

x = a \cos (θ - α) \Rightarrow θ = α - \arccos \frac{x}{a}

y = b \cos (α - β - \arccos \frac{x}{a})

[\cos (u + v) = \cos u \cos v - \sin u \sin v]

[\sin (u + v) = \cos u \sin v - \sin u \cos v]

y = \frac{b}{a} x \cos (α - β) \pm \frac{b}{a} \sqrt{a^{2} - x^{2}} \sin (α - β)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{2 x y}{a b} \cos (α - β) = \sin^{2} (α - β)

\Rightarrow y = \frac{b}{a} x \cos (α - β) \pm \frac{b}{a} \sqrt{a^{2} - x^{2}} \sin (α - β)

same as above \Rightarrow proved

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