Question

Find the equation of an ellipse whose major axis lies on the x-axis and which passes through the points (4, 3) and (6, 2).

Answer 1

Since the major axis of the ellipse lies on the x-axis, so it is a horizontal ellipse.

Let the required equation of the ellipse be

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(where{a}^2>b^2)....\left(i\right)$

Since, (4, 3) lies on (i), we have $\frac{16}{a^2}+\frac{9}{b^2}=1....\left(ii\right)$

Also, since (6, 2) lies on (i), we have $\frac{36}{a^2}+\frac{4}{b^2}=1....\left(iii\right)$

Putting $\frac{1}{a^2}=u$ and $\frac{1}{b^2}=v$ in (ii) and (iii), we get

$16u+9v=1...\left(iv\right)$

$36u+4v=1...\left(v\right)$

On multiplying (iv) by 9 and (v) 4, subtracting, we get

$65v=5\iff v=\frac{1}{13}\iff \frac{1}{b^2}=\frac{1}{13}\iff b^2=13.$

Putting $\frac{v}{13}$ in (iv), we get

$16u=(1-\frac{9}{13})\iff 16u=\frac{4}{13}\iff u=(\frac{4}{13}\times \frac{1}{16})=\frac{1}{52}$

$\iff \frac{1}{a^2}=\frac{1}{52}\iff a^2=52.$

Thus, $a^2=52$ and $b^2=13.$

Hence, the required equation is $\frac{x^2}{52}+\frac{y^2}{13}=1.$