Question

Find the equation of the ellipse with centre at the origin, one vertex at (4,0) and which passes through the point $(2,\frac{\sqrt{3}}{2})$.

Answer 1

We have,

General equation of Ellipse is with centre (h, k) is

$\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$ (When the verticex are horizontally oriented)

The general form for horizontally oriented are:-

$(h-a,k)and(h+a,k)$

The general form and the given vertices $(4,0)$ then,

$h-a=-4$

$h=0$

$then,$

$a=4$

Substitute these value into equation (1).

$\frac{{(x-0)}^2}{4^2}+\frac{{(y-0)}^2}{b^2}=1$

$\frac{x^2}{16}+\frac{y^2}{b^2}=1$

Substitute the point $(2,\frac{\sqrt{3}}{2})$.

$\frac{{\left(2\right)}^2}{4^2}+\frac{{\left(\frac{\sqrt{3}}{2}\right)}^2}{b^2}=1$

$\frac{1}{4}+\frac{3}{4b^2}=1$

$\frac{3}{4b^2}=1-\frac{1}{4}$

$\frac{3}{4b^2}=\frac{3}{4}$

$b^2=1$

$b=1$

Hence, the required equation is ellipse is $\frac{x^2}{16}+\frac{y^2}{1}=1$