Question

If x=a cos 0;y=bsin 0 then eliminate 0​

Answer 1

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

$\rm :\longmapsto\:x = a \: cos\theta$

and

$\rm :\longmapsto\:y \: = \: b \: sin\theta$

$\large\underline{\sf{To\:Find - }}$

$\red{\bf :\longmapsto\:To \: eliminate \: \theta }$

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

Given that,

$\rm :\longmapsto\:x = a \: cos\theta$

can be rewritten as

$\rm :\longmapsto\:cos\theta = \dfrac{x}{a} - - - (1)$

Also given that,

$\rm :\longmapsto\:y \: = \: b \: sin\theta$

can be rewritten as

$\rm :\longmapsto\:sin\theta = \dfrac{y}{b} - - - (2)$

We know, from Trigonometric Identities,

$\red{\rm :\longmapsto\: {sin}^{2}\theta + {cos}^{2}\theta = 1 }$

On substituting the values from equation (1) and equation (2), we get

$\rm :\longmapsto\: {\bigg(\dfrac{x}{a} \bigg) }^{2} + {\bigg(\dfrac{y}{b} \bigg) }^{2} = 1$

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

Hence, is the required result

Additional Information :-

$\rm \: 1) \: x = acos\theta \: and \: y = bsin\theta \: are \: parametric \: coordinates \: of \\ \boxed{ \bf{ \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1}}$

$\rm \: 2) \: x = asec\theta \: and \: y = btan\theta \: are \: parametric \: coordinates \: of \\ \boxed{ \bf{ \: \dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1}}$

$\rm \: 3) \: x = acos\theta \: and \: y = asin\theta \: are \: parametric \: coordinates \: of \\ \boxed{ \bf{ \: {x}^{2} + {y}^{2} = {a}^{2} }}$