Question

✨✨✨✨✨✨✨ any genuis here who can solve this✨✨✨✨✨✨

Answer 1

Hiii. ...friend,

The answer is here,

$let \: theta \: be \: \alpha \: and \: \: phi \: \: be \: \: be \: \beta \: for \: my \: convinience$.Then

$y = bsec \alpha . \sin \beta$.

$x = asec \alpha . \cos \beta$

From the question,

$= > \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }$

$= > \: \frac{ ({ {a}^{2} {sec}^{2} \alpha }. {cos}^{2} \beta ) }{ {a}^{2} } + \frac{ {b}^{2}. {sec}^{2} \alpha . {sin}^{2} \beta }{ {b}^{2} }$

$= > \: {sec}^{2} \alpha . {cos}^{2} \beta + {sec }^{2} \alpha . { \sin}^{2} \beta$

$= > \: {sec}^{2} \alpha ( {cos}^{2} \alpha + {sin}^{2} \alpha )$

$since \: {sin}^{2} \beta + {cos}^{2} \beta = 1$

$= > \: {sec}^{2} \alpha$

$= > \: 1 + {tan}^{2} \alpha .............(1)$

And it is given that,

$= > \: z = ctan \alpha$

$= > \: tan \alpha = \frac{z}{c}$

Substitute this value in eq. (1)

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

This is the answer .

:-) Hope if helps u.

Answer 2

Given, x = a secθcosΦ

⇒ x / a = secθcosΦ

Square on both sides,

⇒ x² / a² = sec²θcos²Φ       ....( i )

Given, y = b secθsinΦ      

⇒ y / b = secθsinΦ

Square on both sides

⇒ y² / b² = sec²θsin²Φ       ....( ii )

Given, z = c tanθ

⇒ z / c = tanθ

Square on both sides,

⇒ z² / c² = tan²θ           ....( iii )

Adding ( i ) & ( ii ) :-

⇒ x² / a² + y² / b² = sec²θcos²Φ + sec²θsin²Φ

⇒ x² / a² + y² / b² = sec²θ( cos²Φ + sin²Φ )

We know, sin²A + cos²A = 1

⇒ x² / a² + y² / b² = sec²θ( 1 )

⇒ x² / a² + y² / b² = sec²θ

We know, sec²A = 1 + tan²A

⇒ x² / a² + y² / b² = 1 + tan²θ

Substituting the value of tan²θ from ( iii )

⇒ x² / a² + y² / b² = 1 + z² / c²

⇒ x² / a² + y² / b² = ( c² + z² ) / c²

Therefore the value of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}\:\:is\:\: \dfrac{c^2 + z^2 }{c^2}$