If ( x a - y b )=1 and ( x a + y b )=1, prove that ( x^2 a^2 + y^2 b^2 )=2.

We have ( x a - y b )=1 Squaring both side, we have: ( x a - y b )^2=(1)^2 ( x^2 a^2 ^2 + y^2 b^2 ^2 -2 x a y b )=1 (i) Again, ( x a + y b )=1 Squaring both side, we get: ( x a + y b )^2=(1)^2 ( x^2 a^2 ^2 + y^2 b^2 ^2 +2 x a y b )=1 (ii) Now, adding (i) and (ii), we get: ( x^2 a^2 ^2 + y^2 b^2 ^2 -2 x a y b ) + ( x^2 a^2 ^2 + y^2 b^2 ^2 +2 x a y b )=2 x^2 a^2 ^2 + y^2 b^2 ^2 - x^2 a^2 ^2 + y^2 b^2 ^2 =2 ( x^2 a^2 ^2 + y^2 a^2 ^2 )+ ( y^2 b^2 ^2 + y^2 b^2 ^2 )=2 x^2 a^2 ( ^2 + ^2 )+ y^2 b^2 ( ^2+ ^2 )=2 x^2 a^2 + y^2 b^2 =2 [ ^2 + ^2 =1 ] x^2 a^2 + y^2 b^2 =2

If ( x a - y b )=1 and ( x a + y b )=1, prove that ( x^2 a^2 + y^…

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Question

If $\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)=1$ and $\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)=1,$ prove that $\Big(\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}\Big)=2.$

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Answer

We have $\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)=1$
Squaring both side, we have:
$\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)^2=(1)^2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=1\dots(\text{i})$
Again, $\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)=1$
Squaring both side, we get:
$\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)^2=(1)^2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta+2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=1\dots(\text{ii})$
Now, adding (i) and (ii), we get:
$\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)\\ \ +\Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta+2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta=2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{a}^2}\cos^2\theta\Big)+\Big(\frac{\text{y}^2}{\text{b}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta\Big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}\big(\sin^2\theta+\cos^2\theta\big)+\frac{\text{y}^2}{\text{b}^2}\big(\cos^2+\sin^2\theta\big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$ $\big[\because\ \sin^2\theta+\cos^2\theta=1\big]$
$\therefore\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$

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