Question

Find the equation of the ellipse in standard form if passing through the points (−3, 1) and (2, −2)

Answer 1

Let the required equation of ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1, where a > b.

The ellipse passes through the points (–3, 1) and (2, –2).

∴ Substituting x = –3 and y = 1 in equation of ellipse, we get

`(-3)^2/"a"^2 + 1^2/"b"^2` = 1

∴ `9/"a"^2 + 1/"b"^2` = 1   ...(i)

Substituting x = 2 and y = –2 in equation of ellipse, we get

`2^2/"a"^2 + (-2)^2/"b"^2` = 1

∴ `4/"a"^2 + 4/"b"^2` = 1   ...(ii)

Let `1/"a"^2` = A and `1/"b"^2` = B

∴ Equations (i) and (ii) become

9A + B = 1 …(iii)

4A + 4B = 1 …(iv)

Multiplying (iii) by 4, we get

36A + 4B = 4 …(v)

Subtracting (iv) from (v), we get

32A = 3

∴ A = `3/32`

Substituting A = `3/32` in (iv), we get

`4(3/32) + 4"B"` = 1

∴ `3/8 + 4"B"` = 1

∴ 4B = `1 - 3/8`

∴ 4B = `5/8`

∴ B = `5/32`

Since `1/"a"^2` = A and `1/"b"^2` = B,

`1/"a"^2 = 3/32` and `1/"b"^2 = 5/32`

∴ a² = `32/3` and b² = `32/5`

∴ The required equation of ellipse is

`x^2/((32/3)) + y^2/((32/5))`, i.e., 3x² + 5y² = 32.