Question

Find the equation of the ellipse, with major axis along the $x$ - axis and passing through the points $(4,3)$ and $(-1,4)$.

Answer 1

Given: Ellipse has major axis along the $x-$axis and passing through the points $(4,3)$ and $(-1,4)$.

$\because$ Major axis is along $x$-axis.

$\therefore$ Required equation of ellipse is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1...\left(1\right)$

$\because$ Points $(4,3)$ & $(-1,4)$ lie on the ellipse

$\therefore$ Points$(4,3)$ & $(-1,4)$ will satisfy equation of ellipse.

From point $(4,3)$ and equation $\left(1\right)$

$\Rightarrow \frac{(4{)}^2}{a^2}+\frac{(3{)}^2}{b^2}=1$

$\Rightarrow \frac{16}{a^2}+\frac{9}{b^2}=1....\left(2\right)$

From point $(-1,4)$ and equation $\left(1\right)$

$\Rightarrow \frac{(-1{)}^2}{a^2}+\frac{(4{)}^2}{b^2}=1$

$\Rightarrow \frac{1}{a^2}+\frac{16}{b^2}=1$

$\Rightarrow \frac{1}{a^2}=1-\frac{16}{b^2}....\left(3\right)$

Now, from equations $\left(2\right)$ and $\left(3\right)$

$\Rightarrow 16(1-\frac{16}{b^2})+\frac{9}{b^2}=1$

$\Rightarrow 16-\frac{256}{b^2}+\frac{9}{b^2}=1$

$\Rightarrow \frac{-256+9}{b^2}=1-16$

$\Rightarrow b^2=\frac{247}{15}$

Putting value of $b^2=\frac{247}{15}$ in equation $\left(3\right)$

i.e. $\frac{1}{a^2}=1-\frac{16}{b^2}$

$\Rightarrow \frac{1}{a^2}=1-\frac{16}{\frac{247}{15}}$

$\Rightarrow \frac{1}{a^2}=1-\frac{16\times 15}{247}$

$\Rightarrow \frac{1}{a^2}=\frac{247-240}{247}$

$\Rightarrow \frac{1}{a^2}=\frac{7}{247}$

$\Rightarrow a^2=\frac{247}{7}$

Thus, $a^2=\frac{247}{7}$ & $b^2=\frac{247}{15}$

Putting values of $a^2$ and $b^2$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Hence, required equation of ellipse is $\frac{x^2}{\left(\frac{247}{7}\right)}+\frac{y^2}{\left(\frac{247}{15}\right)}=1$