Question

xcosA/a+ ysinA /b= 1 and xsinA/a - ycosA/b = 1,prove that x^2/a^2 + y^2/b^2=2

Answer 1

Hey!!!!

xcosA/a + ysinA/b = 1 ------1)

xsinA/a - ycosA /b = 1 -------2)

1st of all adding the both equation and squaring them .

we get

(xcosA/a + ysinA/b )² +(xsinA/a - ycosA/b)² =1²+1²

x²cos²A/a² + y² sin²A/b²+2xcos/a ×ysinA/b + x²sin²A/a² + y²cos²A/b² = 2

x²cos²A /a² + x²sin²A/a² + y2sin²A/b² +y²cos²A/b² =2
[just rearranging the term ]

x²/a² (sin²A+ cos²A) + y²/b²(sin²A+cos²A) = 2

x²/a²×1 + y²/b²×1 = 2

x²/a² + y²/b² = 2 [prooved ]

~~~~~~~~~~~~~~~~~~~~~~~~

Hope it helps you !!!!

@Rajukumar111

Answer 2

Solution :

_____________________________________________________________

Given :

⇒ $\frac{x cos A}{a} + \frac{y sin A }{b} = 1$ ...(i)

⇒ $\frac{xsinA}{a} - \frac{ycosA}{b} = 1$ ....(ii)

_____________________________________________________________

To Prove :

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



_____________________________________________________________

Proof :

Adding (i)² & (ii)²,

We get,

$(\frac{x cos A}{a} + \frac{y sin A }{b})^2 + (\frac{xsinA}{a} - \frac{ycosA}{b})^2 = 1^2 + 1^2$

⇒ [tex] (\frac{xcos A}{a})^2 + (\frac{ysinA}{b})^2 + \frac{2(xcosA)(ysinA)}{(a)(b)} + (\frac{xsinA}{a})^2 + (\frac{ycosA}{b})^2 - \frac{2(xsinA)(ycosA)}{(a)(b)} = 1 + 1[/tex]

⇒ $\frac{x^2cos^2 A}{a^2} + \frac{y^2sin^2A}{b^2} + \frac{2xysinAcosA}{ab} + \frac{x^2sin^2A}{a^2} + \frac{y^2cos^2A}{b^2} - \frac{2xysinAcosA}{ab} = 2$

⇒ $\frac{x^2cos^2A}{a^2} \frac{x^2sin^2A}{a^2} + \frac{y^2sin^2A}{b^2} +\frac{y^2cos^2A}{b^2} = 2$

⇒ $\frac{x^2}{a^2} (sin^2A + cos^2A) + ( \frac{y^2}{b^2})(sin^2A + cos^2A) = 2$

⇒ $( \frac{x^2}{a^2} + \frac{y^2}{b^2} )(sin^2A + cos^2A) = 2$


We know that,

sin² A + cos² A = 1,.

Hence,

⇒ $( \frac{x^2}{a^2} + \frac{y^2}{b^2} )(1) = 2$

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

= RHS,.

              ∴ Hence, proved,.

_____________________________________________________________

                                    Hope it Helps!!