Question

If x = a sec θ cos ϕ, y = b θ sin ϕ and z = c tan θ, then $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=?$
(a) $\left(1+\frac{z^2}{c^2}\right)$
(b) $\left(1-\frac{z^2}{c^2}\right)$
(c) $\left(\frac{z^2}{c^2}-1\right)$
(d) $\frac{z^2}{c^2}$

Answer 1

(a) $\left(1+\frac{z^2}{c^2}\right)$

$$\begin{gathered} x=a\text{sec}\theta \text{cos}\varphi ,y=b\text{sec}\theta \text{sin}\varphi \mathrm{and}z=\text{ctan}\theta \\ \therefore \text{sec}\theta \text{cos}\varphi \text{=}\frac{x}{a}\text{, sec}\theta \text{sin}\varphi =\frac{y}{b}\text{and tan}\theta \text{=}\frac{z}{c} \\ {\text{Now, sec}}^2\theta {\text{cos}}^2\varphi +{\text{sec}}^2\theta {\text{sin}}^2\varphi ={\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2\left[\text{Squaring both sides}\right] \\ {\text{=>sec}}^2\theta ({\text{cos}}^2\varphi +{\text{sin}}^2\varphi )=\frac{x^2}{a^2}+\frac{y^2}{b^2} \\ {\text{=>sec}}^2\theta =\frac{x^2}{a^2}+\frac{y^2}{b^2} \\ \text{=>1}+{\text{tan}}^2\theta =\frac{x^2}{a^2}+\frac{y^2}{b^2} \\ \text{=>1+}\frac{z^2}{c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2} \\ \therefore (\frac{x^2}{a^2}+\frac{y^2}{b^2})=\left(\text{1+}\frac{z^2}{c^2}\right) \end{gathered}$$