Question

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

Answer 1

`\displaystyle \frac{ {3x}^(2) }{ 35 } + \frac{{2y}^(2) }{ 35 } = 1`

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 = (3)/(35) \\ \\ p = (2)/(35) \end{cases}`

substitute back:

`\displaystyle \begin{cases} \frac{1}{ {a}^(2) } = (3)/(35) \\ \\ \frac{1}{ {b}^(2) } = (2)/(35) \end{cases}`

divide both equation by 1 which yields:

`\displaystyle \begin{cases} {a}^(2) = (35)/( 3) \\ \\ {b}^(2) = (35)/(2) \end{cases}`

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

`\displaystyle \frac{ {x}^(2) }{ (35)/(3) } + \frac{{y}^(2) }{ (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)