Question

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
through the points (-3, 1) and (2,-2) is
(A) 5x2 + 3y =32 (B) 3r? + 5y = 32 (C) 5x° - 3y2 = 32 (D) 3x* + 5y? + 32 = 0​

Answer 1

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,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

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,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

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,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

Hence equation of the ellipse is 3x² + 5y² = 32