Question

12. Find the equation of ellipse if it passes through 3,2
whose centre (0,0) eccentricity root 3 by 2
and the major axis

on y-axis.​

Answer 1

Answer:

$\Large \text { \frac{y^{2} }{40 } \: + \: \frac{x^{2} }{10} \: = \: 1 }$

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Step-by-step explanation:

TO KNOW  :-

• The general eqn. of ellipse with center (0,0) and major axis on y-axis :-

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

• If a  point lies on the ellipse then it should satisfy the equation of the ellipse.

• Eccentricity of an ellipse (e) =  $\sqrt{1\: - \: \frac{b^{2}}{a^{2}} }$

POINTS TO NOTE ABOUT THIS ELLIPSE :-

• Center of the ellipse = (0, 0)

• Eccentricity (e) = $\sqrt{3}$/ 2

• Major axis on y-axis
• Passes through the point (3, 2)

SOLUTION :-

Since, the major axis of the ellipse lies on the Y-Axis.

The general eqn. of ellipse with center (0,0) :-

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

Here,

1. a = length of semi-major axis
2. b = length of semi-minor axis

Let, the point (3, 2) be P.

• Now, if the point P lies on the ellipse then it should satisfy the equation of the ellipse. Therefore :-

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

=> $\: \frac{2^{2} }{a^{2} } \: + \: \frac{3^{2} }{b^{2} } \: = \: 1$

=> $\frac{4b^{2} \:+ \: 9a^{2}}{a^{2}b^{2}} =1$

=> 4b² + 9a² = a²b² . . . . . . . . . . (i)

• Eccentricity of an ellipse (e) =  $\sqrt{1\: - \: \frac{b^{2}}{a^{2}} }$

=> $e^{2} = \frac{a^{2} \: - \: b^{2}}{a^{2}}$

=> $(\frac{\sqrt{{3} } }{2} )^{2} = \frac{a^{2}\: - \: b^{2}}{a^{2}}$

=> $\frac{3 }{4} = \frac{a^{2}\: - \: b^{2}}{a^{2}}$

=> 3a² = 4a² - 4b²

=> 4b² = a² . . . . . . . . . . . . . . . (ii)

• Evaluating eqn. (i) and (ii) :-

=> 4b² + 9(4b²) = (4b²)b²   [substituting the value of a² from eqn (i) in eqn (ii)]

{Cancelling 4b² from both the sides} -

=> 1 +9 = b²

=> 10 = b²

from eqn (i), 4b² = a²

∴ => 4 × 10 = a²

=> 40 = a²

Hence, the equation of the ellipse is :-

=> $\boxed{\Large \text { \frac{y^{2} }{40 } \: + \: \frac{x^{2} }{10} \: = \: 1 } }$