If x = a secθ cosΦ, y = b secθ sinΦ, z = c tan θ, then x2/a2 +�

1.	If x = a secθ cosΦ, y = b secθ sinΦ, z = c tan θ, then x2/a2 + y2/b2 = ………A) z2/c2B) 1 – z2/c2C) z2/c2 – 1D) 1 + z2/c2
Answer» Correct option is: D) \(1 + \frac{z^2}{c^2}\) We have x = a sec \(\theta\) cos \(\phi\) = \(\frac xa\) sec \(\theta\) cos \(\phi\). y = b sec \(\theta\) sin \(\phi\) = \(\frac yb\) = sec \(\theta\) sin \(\phi\). z = c tan \(\theta\) Now, \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = (\frac xa)^2 + (\frac yb)^2\) = \((sec\, \theta \, cosc \, \phi)^2 + (sec\, \theta \, sin\, \phi)^2\) = \(sec^2\theta \, cos^2\phi + sec^2\theta \, sin^2\phi\) = \(sec^2\theta (cos^2\, \phi + sin^2\phi)\) = \(sec^2\theta \) (\(\because\) \(cos^2 \phi + sin^2\phi = 1\)) = 1 + \(tan^2\theta\) (\(\because\) \(sec^2\theta = 1+ tan^2\theta\)) = 1+ \((\frac zc)^2\) (\(\because\) z = c \(\tan \theta = tan \theta = \frac zc\)) = 1 + \(\frac {z^2}{c^2}\) Correct option is: D) 1 + \(\frac{z^2}{c^2}\)

If x = a secθ cosΦ, y = b secθ sinΦ, z = c tan θ, then x2/a2 + y2/b2 = ………A) z2/c2B) 1 – z2/c2C) z2/c2 – 1D) 1 + z2/c2

Answer»

Correct option is: D) \(1 + \frac{z^2}{c^2}\)

We have

x = a sec \(\theta\) cos \(\phi\) = \(\frac xa\) sec \(\theta\) cos \(\phi\).

y = b sec \(\theta\) sin \(\phi\) = \(\frac yb\) = sec \(\theta\) sin \(\phi\).

z = c tan \(\theta\)

Now, \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = (\frac xa)^2 + (\frac yb)^2\)

= \((sec\, \theta \, cosc \, \phi)^2 + (sec\, \theta \, sin\, \phi)^2\)

= \(sec^2\theta \, cos^2\phi + sec^2\theta \, sin^2\phi\)

= \(sec^2\theta (cos^2\, \phi + sin^2\phi)\)

= \(sec^2\theta \) (\(\because\) \(cos^2 \phi + sin^2\phi = 1\))

= 1 + \(tan^2\theta\) (\(\because\) \(sec^2\theta = 1+ tan^2\theta\))

= 1+ \((\frac zc)^2\) (\(\because\) z = c \(\tan \theta = tan \theta = \frac zc\))

= 1 + \(\frac {z^2}{c^2}\)

Correct option is: D) 1 + \(\frac{z^2}{c^2}\)