Question

Find the equation of the ellipse, with centre at the origin, major axis on the Y-axis and passing through the point (3,2) and (1,6) ​

Answer 1

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

$\sf \: Let \: the \: equation \: of \: ellipse \: be \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 - - (1)$

Now,

It is given that

• Equation (1) passes through (3, 2),

Therefore,

$\rm :\longmapsto\:\dfrac{ {3}^{2} }{ {a}^{2} } + \dfrac{ {2}^{2} }{ {b}^{2} } = 1$

$\rm :\longmapsto\:\dfrac{9}{ {a}^{2} } + \dfrac{4}{ {b}^{2} } = 1 - - - (2)$

Also,

• Equation (1) passes through (1, 6),

Therefore,

$\rm :\longmapsto\:\dfrac{ {1}^{2} }{ {a}^{2} } + \dfrac{ {6}^{2} }{ {b}^{2} } = 1$

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } + \dfrac{ 36}{ {b}^{2} } = 1 - - - (3)$

Now, we have to solve equation (2) and equation (3) to get the values of a and b,

Multiply equation (3) by 9, we get

$\rm :\longmapsto\:\dfrac{9}{ {a}^{2} } + \dfrac{ 324}{ {b}^{2} } = 9 - - - (4)$

On Subtracting, equation (3) fom equation (4), we get

$\rm :\longmapsto\:\dfrac{320}{ {b}^{2} } = 8$

$\bf\implies \: {b}^{2} = 40 - - (5)$

Substituting the values of equation (5) in equation (3), we get

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } + \dfrac{ 36}{40 } = 1$

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } + \dfrac{ 9}{10 } = 1$

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } = 1 - \dfrac{ 9}{10 }$

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } = \dfrac{ 10 - 9}{10 }$

$\rm :\longmapsto\:\dfrac{ 1 }{ {a}^{2} } = \dfrac{ 1}{10 }$

$\bf\implies \: {a}^{2} \: = \: 10 \: - - - (6)$

On substituting (6) and (5) in equation (1), we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{10} + \dfrac{ {y}^{2} }{40} = 1$

$\bf\implies \: {4x}^{2} + {y}^{2} = 40$

Additional Information :-

$\sf \: For \: the \: ellipse \: : \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

$\boxed{ \bf{Vertex \: \: = \: \: ( \pm \: a, \: 0)}}$

$\boxed{ \bf{eccentricity \: (e) = \sqrt{1 - \dfrac{ {b}^{2} }{ {a}^{2} } } }}$

$\boxed{ \bf{Focus \: = \: ( \pm \: ae, \: 0)}}$

$\boxed{ \bf{Length \: of Latus \: Rectum \: = \dfrac{ {2b}^{2} }{a} }}$

$\boxed{ \bf{Length \: of \: major \: axis \: = 2a}}$

$\boxed{ \bf{Length \: of \: minor \: axis \: = 2b}}$