Question

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - x^2/c^2 = 1`

Answer 1

Given:

`x = a sec theta cos phi`

`=> x/a = sec theta  cos phi`   ........(1)

`y = b sec theta sin phi`

`=> y/b = sec theta sin phi`

`=> y/b = sec theta sin phi`

`=> zx/c =  tan theta`

We have to prove that `x^2/a^2  + y^2/b^2 - z^2/c^2  = 1`

Squaring the above equations and then subtracting the third from the sum of the first two, we have

`(x/a)^2 + (y/b)^2  - (z/c)^2 = (sec theta cos phi)^2 + (sec theta sin phi)^2 - (tan theta)^2`

`=> x^2/ a^2 + y^2/b^2 - z^2/c62 = sec^2 theta cos^2 phi + sec^2 theta sin^2 phi - tan^2 theta` 

`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = (sec^2 theta cos^2 phi + sec^2 theta sin&2 phi) - tan^2 theta`

`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = sec^2 theta(cos^2 phi + sin^2 phi) - tan^2 theta`

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

`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = sec^2 theta - tan^2 theta`

`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`

Hence proved.