Question

If x/a cos θ+ y/ b sin θ =1 and x/a cos θ-y/b sin θ= -1 prove that x²/a² + y²/b² = 2..​

Answer 1

Given :-

$\tt \: \frac{x}{a} \: sin \theta \: - \: \frac{y}{b} \: cos \theta \: = 1 \: \:$

$\tt \frac{x}{a} \: sin \theta \: + \: \frac{y}{b} \: cos \theta \: = -1$

To prove :-

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

Proof :-

$\implies \tt \: \frac{x}{a} \: sin \theta \: - \: \frac{y}{b} \: cos \theta \: = 1 \: \:$

On squaring both the sides,

$\implies \tt \: {\huge{(}}{\frac{x}{a} \: sin \theta \: - \: \frac{y}{b} \: cos \theta{ \huge{) {}^{2} }} \: = (1 ) {}^{2} } \: \:$

We know that,

(a-b)² = a² + b² - 2ab

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: - \: 2. \frac{x}{a} \: sin \theta. \frac{y}{b} \: cos \theta \: = 1 \: \:$

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: - \: \frac{2xy}{ab} \: sin \theta. \: cos \theta \: = 1 \: \:$

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: = 1 + \frac{2xy}{ab} \: sin \theta. \: cos \theta \: \:$

Taking x²/a² + y²/b² common on L.H.S and taking L.C.M on R.H.S,

$\implies \tt \: { \huge{(}}\frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }{ \huge{)}} ( \: sin {}^{2} \theta \: + \: {cos}^{ 2} \theta)\: = \frac{ab + 2xy \: sin \theta. \: cos \theta }{ab} \: \:$

We know that,

(sin²θ + cos²θ) = 1

$\implies \tt \: { \huge{(}}\frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }{ \huge{)}} ( 1)\: = \frac{ab + 2xy \: sin \theta. \: cos \theta }{ab} \: \:$

$: \implies { \green{\tt\: \frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }\: = \frac{ab + 2xy \: sin \theta. \: cos \theta }{ab} \: \: }}$

Now,

It was also given that,

$: \implies\tt \frac{x}{a} \: sin \theta \: + \: \frac{y}{b} \: cos \theta \: = -1$

On squaring both the sides,

$\implies \tt \: {\huge{(}}{\frac{x}{a} \: sin \theta \: + \: \frac{y}{b} \: cos \theta{ \huge{) {}^{2} }} \: = (-1 ) {}^{2} } \: \:$

We know that,

(a+b)² = a² + b² + 2ab

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: + \: 2. \frac{x}{a} \: sin \theta. \frac{y}{b} \: cos \theta \: = 1 \: \:$

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: + \: \frac{2xy}{ab} \: sin \theta. \: cos \theta \: = 1 \: \:$

$\implies \tt \: \frac{x {}^{2} }{a {}^{2} } \: sin {}^{2} \theta \: + \: \frac{ {y}^{2} }{ {b}^{2} } {cos}^{2} \theta\: = 1 - \frac{2xy}{ab} \: sin \theta. \: cos \theta \: \:$

Again, taking x²/a² + y²/b² common on L.H.S and taking L.C.M on R.H.S,

$\implies \tt \: { \huge{(}}\frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }{ \huge{)}} ( \: sin {}^{2} \theta \: + \: {cos}^{ 2} \theta)\: = \frac{ab - 2xy \: sin \theta. \: cos \theta }{ab} \: \:$

We know that,

(sin²θ + cos²θ) = 1

$\implies \tt \: { \huge{(}}\frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }{ \huge{)}} ( 1)\: = \frac{ab - 2xy \: sin \theta. \: cos \theta }{ab} \: \:$

$: \implies { \green{\tt\: \frac{x {}^{2} }{a {}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }\: = \frac{ab - 2xy \: sin \theta. \: cos \theta }{ab} \: \: }}$

So,

We got the values for both given eq.

Now,

We need to prove,

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

Put the values of x²/a² + y²/b² that we got above,

L.H.S =>

$\implies \tt \: \frac{ab + 2xy \: sin \theta.cos \theta}{ab } + \frac{ab - 2xy \: sin \theta.cos \theta}{ab }$

$\implies \tt \: \frac{ab + 2xy \: sin \theta.cos \theta +ab - 2xy \: sin \theta.cos \theta }{ab }$

$\implies \tt \: \frac{ab + \cancel{2xy \: sin \theta.cos \theta }+ab - \cancel{2xy \: sin \theta.cos \theta }}{ab }$

$\implies \tt \: \frac{2 \: \cancel{ab}}{ \cancel{ab}} \: = 2{ \red{ \: = R.H.S }} \: \: (proved)$