Answer:
Equation of the ellipse = 3x² + 5y² = 32
Step-by-step explanation:
Given:
- The centre of the ellipse is at the origin and the X axis is the major axis
- It passes through the points (-3, 1) and (2, -2)
To Find:
- The equation of the ellipse
Solution:
The equation of an ellipse is given by,
[tex]\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1[/tex]
Given that the ellipse passes through the point (-3, 1)
Hence,
[tex]\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1[/tex]
Cross multiplying we get,
- 9b² + a² = 1 ²× a²b²
- a²b² = 9b² + a²
Multiply by 4 on both sides,
- 4a²b² = 36b² + 4a²------(1)
Also by given the ellipse passes through the point (2, -2)
Substituting this,
[tex]\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1[/tex]
Cross multiply,
- 4b² + 4a² = 1 × a²b²
- a²b² = 4b² + 4a²-------(2)
Subtracting equations 2 and 1,
- 3a²b² = 32b²
- 3a² = 32
- a² = 32/3----(3)
Substituting in 2,
- 32/3 × b² = 4b² + 4 × 32/3
- 32/3 b² = 4b² + 128/3
- 32/3 b² = (12b² + 128)/3
- 32b² = 12b² + 128
- 20b² = 128
- b² = 128/20 = 32/5
Substituting the values in the equation for ellipse,
[tex]\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1[/tex]
[tex]\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1[/tex]
Multiplying whole equation by 32 we get,
3x² + 5y² = 32
Hence equation of the ellipse is 3x² + 5y² = 32
