Question

If x = a sin α, cos β, y = b sin α.sin β and z = c cos α then (x²/a²) + (y²/b²) + (z²/c²) = ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given:

$\rm \longmapsto x = a \sin \alpha \cos \beta$

$\rm \longmapsto y = b \sin \alpha \sin \beta$

$\rm \longmapsto z = c\cos \alpha$

Therefore:

$\rm \longmapsto \dfrac{x}{a} = \sin \alpha \cos \beta$

$\rm \longmapsto \dfrac{y}{b} = \sin \alpha \sin \beta$

$\rm \longmapsto \dfrac{z}{c} = \cos \alpha$

Now:

$\rm = \dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }$

$\rm = { \sin}^{2} \alpha \cos^{2} \beta + { \sin}^{2} \alpha \sin^{2} \beta + { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha (\cos^{2} \beta + \sin^{2} \beta )+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha \cdot1+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha + { \cos}^{2} \alpha$

$\rm = 1$

Therefore:

$\rm \longmapsto\dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} } = 1$