Question

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]

पर्याय

• \[\frac{z^2}{c^2}\]

• \[1 - \frac{z^2}{c^2}\]

• \[\frac{z^2}{c^2} - 1\]

• \[1 + \frac{z^2}{c^2}\]

Answer 1

Given: 

`x= a secθcosΦ` 

`⇒ x/a=secθ cosΦ `

`y=b sec θ sinΦ `

`⇒ y/b=secθ sinΦ `

`z=c tan θ`

`z/c= tan θ` 

Now, 

`(x/a)^2+(y/b)^2-(z/c)^2=(secθ cosΦ)^2+(secθ sin Φ)^2-(tanθ )^2` 

`⇒ x^2/a^2+y^2/b^2-z^2/c^2= sec^2θcos^2 Φ+sec^2θsin^2Φ-tan^2θ`

`⇒ x^2/a^2+y^2/b^2-z^2/c^2=(sec^2θ cos^2Φ+sec^2θ sin^2 sin^2Φ)-tan^2Φ`

`⇒ x^2/a^2+y^2/b^2-z^2/c^2=sec^2θ(cos^2Φ+sin^2Φ)-tan^2θ`

`⇒ x^2/a^2+y^2/b^2-z^2/c^2= sec^2θ(1)-tan^2θ`

`⇒ x^2/a^2+y^2/b^2-z^2/c^2=sec^2θ-tan^2θ`

`⇒ x^2/a^2+y^2/b^2-z^2/c^2=1` 

`⇒x^2/a^2+y^2/b^2=1+z^2/c^2`