Question

6 () Find the points
2V6 units from the point (1, - 2, 3).
IS
Find th
on the X-axis, which are at a distance
from (1, 2, 3)​

Answer 1

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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