Question

If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then x²/a²+y²/b²
A. z²/c²
B. 1-z²/c²
C. z²/c²-1
D.1+z²/c²

Answer 1

x²/a² + y²/b² = z²/c² + 1

Step-by-step explanation:

Given, x= asecθ.cosϕ

x/a = secθ.cosϕ

Squaring on both sides, x²/a² = sec²θ.cos²ϕ -------------(1)

Given, y = bsecθ.sinϕ

y/b = secθ.sinϕ

Squaring on both sides, y²/b² = sec²θ.sin²ϕ ----------------(2)

Given, z = ctanθ

squaring on both sides, z² = c²tan²θ -----------(3)

We know that cos²ϕ + sin²ϕ = 1

From (1) and (2), we get: x²/a² + y²/b² = sec²θ.cos²ϕ + sec²θ.sin²ϕ

= sec²θ (cos²ϕ + sin²ϕ)

 = sec²θ

x²/a² + y²/b² = tan²θ + 1 -----------------(4)

From (3) and (4), we get: x²/a² + y²/b² = tan²θ + 1

x²/a² + y²/b² = z²/c² + 1

Hence proved.

Option D is the answer.