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

Answer 2

Correct Question:-

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 + 3y² =32

(B) 3x² + 5y² = 32

(C) 5x² - 3y² = 32

(D) 3x² + 5y² + 32 = 0

Given:-

• The centre of the ellipse is at the origin and the x-axis the major axis.
• It passes through the points (-3, 1) and (2,-2)

To Find:-

• Equation of the ellipse = ?

Solution:-

As we know,

Standard equation of ellipse is :

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

• We know that it passes through (-3,1) and (2,-2)

When it passes through (-3,3)

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

$\sf \: 9 {b}^{2} + {a}^{2} = {a}^{2} {b}^{2} \: \: \: ...(1)$

When it passes through (2,-2)

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

$\sf \: 4 {a}^{2} + 4 {b}^{2} = {a}^{2} {b}^{2} \: \: \: ...(2)$

$\sf \: equation \: (1) = equation(2)$

$= \sf9{b}^{2} + {a}^{2} = 4 {a}^{2} + 4 {b}^{2}$

$= \sf \: 3 {a}^{2} = 5{b}^{2}$

$= \sf \: {b}^{2} = \frac{3}{5} {a}^{2} \: \: \: ...(3)$

Now,

Equation becomes :

$\sf \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ 5{y}^{2} }{3 {a}^{2} } = 1$

$\sf \frac{3 {x}^{2} + 5 {y}^{2} }{3 {a}^{2} } = 1$

$\sf \: 3 {x}^{2} + 5 {y}^{2} = 3 {a}^{2} \: \: \: ...(4)$

when it passes through (-3,1)

$\sf3 \times ( - 3)^{2} + 5 \times 1 = 3 {a}^{2}$

$\sf27 + 5 = 3{a}^{2}$

$\sf3 {a}^{2} = 32 \: \: \: ...(5)$

Putting the value of 3a² in equation 4

we get

• 3x² + 5y² = 32

Hence,Option B is correct

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