Question

Find the equation of ellipse passing throgh (1,4) and (-3,2)​

Answer 1

Answer:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

Step-by-step explanation:

we want to figure out the ellipse equation which passes through (1,4) and (-3,2)

the standard form of ellipse equation is given by:

$\displaystyle \frac{(x - h {)}^{2} }{ {a}^{2} } + \frac{(y - k {)}^{2} }{ {b}^{2} } = 1$

where:

• (h,k) is the centre
• a is the horizontal redius
• b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

$\displaystyle \frac{ {x}^{2} }{ {a}^{2} } + \frac{{y}^{2} }{ {b}^{2} } = 1$

substituting the value of x and y from the point (1,4),we'd acquire:

$\displaystyle \frac{ 1}{ {a}^{2} } + \frac{16}{ {b}^{2} } = 1$

similarly using the point (-3,2), we'd obtain:

$\displaystyle \frac{ 9}{ {a}^{2} } + \frac{4 }{ {b}^{2} } = 1$

let 1/a² and 1/b² be q and p respectively and transform the equation:

$\displaystyle \begin{cases} q + 16p = 1 \\ 9q + 4p = 1 \end{cases}$

solving the system of linear equation will yield:

$\displaystyle \begin{cases} q = \dfrac{3}{35} \\ \\ p = \dfrac{2}{35} \end{cases}$

substitute back:

$\displaystyle \begin{cases} \dfrac{1}{ {a}^{2} } = \dfrac{3}{35} \\ \\ \dfrac{1}{ {b}^{2} } = \dfrac{2}{35} \end{cases}$

divide both equation by 1 which yields:

$\displaystyle \begin{cases} {a}^{2} = \dfrac{35}{ 3} \\ \\ {b}^{2} = \dfrac{35}{2} \end{cases}$

substitute the value of a² and b² in the ellipse equation , thus:

$\displaystyle \frac{ {x}^{2} }{ \dfrac{35}{3} } + \frac{{y}^{2} }{ \dfrac{35}{2} } = 1$

simplify complex fraction:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

and we're done!

(refer the attachment as well)