If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then x2a2+y2b2=

Solution:

Given:

$x=a \sec \theta \cos \phi$

$\Rightarrow \frac{x}{a}=\sec \theta \cos \phi$

$y=b \sec \theta \sin \phi$

$\Rightarrow \frac{y}{h}=\sec \theta \sin \phi$

$z=c \tan \theta$

$\Rightarrow \frac{z}{c}=\tan \theta$

Now,

$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}-\left(\frac{z}{c}\right)^{2}=(\sec \theta \cos \phi)^{2}+(\sec \theta \sin \phi)^{2}-(\tan \theta)^{2}$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\sec ^{2} \theta \cos ^{2} \phi+\sec ^{2} \theta \sin ^{2} \phi-\tan ^{2} \theta$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\left(\sec ^{2} \theta \cos ^{2} \phi+\sec ^{2} \theta \sin ^{2} \phi\right)-\tan ^{2} \theta$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)-\tan ^{2} \theta$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\sec ^{2} \theta(1)-\tan ^{2} \theta$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\sec ^{2} \theta-\tan ^{2} \theta$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1$

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1+\frac{z^{2}}{c^{2}}$

Hence, the correct option is (d).