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 \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\), 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

\(\frac {-3^2}{a^2} + \frac {1^2}{b^2} = 1\)

∴\(\frac {9}{a^2} + \frac {1}{b^2} = 1\)…..(i)

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

\(\frac {2^2}{a^2} + \frac {(-2)^2}{b^2} = 1\)

∴\(\frac {4}{a^2} + \frac {4}{b^2} = 1\)……(ii)

Let 1/a² = A and 1/b² = B

∴ Equations (i) and (ii) become 

9A + B = 1 ..…(iii) 

4A + 4B = 1 …..(iv) 

Multiplying (iii) by 4, we get 3

6A + 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) + 4B = 1

∴ 3/8 + 4B =1

∴ 4B = 1 – 3/8

∴ 4B = 5/8

∴ B = 5/32

Since 1/a² = A and 1/b² =B

1/a² = 3/32 and 1/b² 5/32

∴ a² = 32/3 and b² = 32/5

∴ The required equation of ellipse is

i.e., 3x² + 5y² = 32.