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Question 141
If x = α secα cosβ, y = b secα sinβ, z = c tan α, then the value of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}$$
Solution
From question
$$x^2 = α^2 sec^2α cos^2β$$
$$\frac{x^2}{a^2} = sec^2α cos^2β$$
$$y^2 = b^2 sec^2α sin^2β $$
$$\frac{y^2}{b^2} = sec^2α sin^2β$$
$$z^2 = c^2 sec^2α tan^2α $$
$$\frac{z^2}{c^2} = sec^2α tan^2α$$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} $$= $$sec^2α cos^2β + sec^2α sin^2β - sec^2α tan^2α$$
$$ sec^2 α - tan^2α =1$$
Hence Option C is the correct answer.
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