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Find the equation for the ellipse that satisfies the given conditions : Major axis on the x-axis and passes through the points $(4,3)$ and $(6,2)$.

$\begin {array} {1 1} (A)\;x^2-4y^2=52 & \quad (B)\;4y^2-x^2=52 \\ (C)\;x^2+4y^2=52 & \quad (D)\;4y^2+x^2=52 \end {array}$

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  • 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 major axis is on the x - axis.
Hence the equation of the ellipse should be of the form $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
It is also given that the ellipse passes through the points (4, 3) and (6, 2).
Now substituting the values for x and y we get,
$ \large\frac{16}{a^2}$$+\large\frac{9}{b^2}$$=1$
$ \Rightarrow 16b^2 +9a^2=a^2b^2$--------(1)
and $\large\frac{36}{a^2}$$+\large\frac{4}{b^2}$$=1$
$ \Rightarrow 36b^2+4a^2=a^2b^2$---------(2)
Let us solve the above equation to obtain $a^2$ and $b^2$
$16b^2+9a^2=a^2b^2 ( \times 4)$
$ 36b^2+4a^2=a^2b^2 ( \times 9)$
$64b^2+ \not{36a}^2 = 4a^2b^2$
$ 324b^2+\not{36a}^2=9a^2b^2$
$(-) \quad (-) \qquad (-)$
$ -260b^2 \qquad = -5a^2b^2$
$ \Rightarrow a^2 = 52 $ and $b^2=13$
Hence the equation of the ellipse is
$ \large\frac{x^2}{52}$$+\large\frac{y^2}{13}$$=1 $
$ \Rightarrow x^2+4y^2=52$
answered Jul 18, 2014 by thanvigandhi_1

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