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If major axis on the $x$-axis and passes through the points $(4,3)$ and $(6,2)$, then find the equation for the ellipse that satisfies the given condition.
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Verified Answer
Major axis is $x$-axis
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(4,3)$ and $(6,2)$ lies on it $\frac{16}{a^2}+\frac{9}{b^2}=1 \quad \ldots$ (i)
$\frac{36}{a^2}+\frac{4}{b^2}=1 \quad \ldots(ii)$
Subtracting $\frac{-20}{a^2}+\frac{5}{b^2}=0 \Rightarrow 5 a^2=20 b^2$ $\Rightarrow a^2=4 b^2$
Putting the value of $a^2$ in eqn. (i) $\frac{16}{4 b^2}+\frac{9}{b^2}=1$ $\therefore b^2=13$ and $a^2=4 b^2=4 \times 13=52$
$\therefore$ Equation of ellipse is $\frac{x^2}{52}+\frac{y^2}{13}=1$
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(4,3)$ and $(6,2)$ lies on it $\frac{16}{a^2}+\frac{9}{b^2}=1 \quad \ldots$ (i)
$\frac{36}{a^2}+\frac{4}{b^2}=1 \quad \ldots(ii)$
Subtracting $\frac{-20}{a^2}+\frac{5}{b^2}=0 \Rightarrow 5 a^2=20 b^2$ $\Rightarrow a^2=4 b^2$
Putting the value of $a^2$ in eqn. (i) $\frac{16}{4 b^2}+\frac{9}{b^2}=1$ $\therefore b^2=13$ and $a^2=4 b^2=4 \times 13=52$
$\therefore$ Equation of ellipse is $\frac{x^2}{52}+\frac{y^2}{13}=1$
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