# How do you write an equation of an ellipse in standard form with foci (8, 0) and (-8, 0) if the minor axis has y-intercepts of 2 and -2?

Jun 1, 2018

Use x- and y-intercepts to deduce standard form ellipse coefficients. Deduce x-intercepts from distance from the foci to known points.

#### Explanation:

We can see from the symmetry of the given points that this ellipse is centred at $\left(x , y\right) = \left(0 , 0\right)$. So the standard form of the ellipse equation is ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$.

When $x = 0$, we know that $y = \pm 2$, so ${y}^{2} = 4$, which gives us ${b}^{2} = 4$.

A definition of an ellipse is the set of points the sum of whose distances from the two foci is a constant. We can from the given information deduce that distance sum - the distance from both foci to either y-intercept is $\sqrt{{8}^{2} + {2}^{2}} = \sqrt{68}$, so the sum is twice that, $2 \sqrt{68}$.

Let the positive x-intercept be at $x = {x}_{i}$. The sum of the distances of this from the foci is ${x}_{i} + 8 + {x}_{i} - 8 = 2 {x}_{i}$, which is equal to $2 \sqrt{68}$, so ${x}_{i} = \sqrt{68}$. At these points $y = 0$, so we deduce that $\frac{68}{a} ^ 2 = 1$, i.e. ${a}^{2} = 68$.

Thus
${x}^{2} / 68 + {y}^{2} / 4 = 1$.