# Find the equation for the ellipse that satisfies the given conditions : Centre at $(0,0),$ major axis on the y-axis and passes through the points $(3, 2)$ and $(1, 6)$

$\begin {array} {1 1} (A)\;x^2+4y^2=40 & \quad (B)\;4x^2-y^2=40 \\ (C)\;4x^2+y^2=40 & \quad (D)\;x^2-4y^2=40 \end {array}$

Toolbox:
• Length of the major axis is 2a
• Length of the minor axis is 2b
• Equation of an ellipse whose major axis is along x - axis is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
• Equation of an ellipse whose minor axis is along y - axis is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
• $c = \sqrt{a^2-b^2}$ whose $c$ is the foci.
Step 1 :
It is given that the centre is at (0,0) and the major axis is on the y - axis.
Hence the equation of the ellipse will be of the form $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
The ellipse passes through the points (3, 2) and (1, 6)
Substituting these values for x and y we get,
Step 2 :
$\large\frac{9}{b^2}$$+\large\frac{4}{a^2}$$=1$
$\Rightarrow 9a^2+4b^2=a^2b^2$--------(1)
and $\large\frac{1}{b^2}$$+ \large\frac{36}{a^2}$$=1$
$a^2+36b^2=a^2b^2$------(2)
Let us solve equation (1) and (2) to obtain the values for $a^2$ and $b^2$
$9a^2+4b^2=a^2b^2 (\times 9)$
$\: a^2+36b^2=a^2b^2$
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$81a^2+36b^2=9a^2b^2$
$\: \: a^2+36b^2=a^2b^2$
$(-) \quad (-) \quad (-)$
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$80a^2 \qquad = 8a^2b^2$
$\therefore b^2=10$
and $a^2=40$
Hence the equation of the ellipse is
$\large\frac{x^2}{10}$$+\large\frac{y^2}{40}$$=1$
or $4x^2+y^2=40$