Question

Which is equation of an ellipse of an ellipse with directrices at x=4 and foci at (2,0) (-2,0)

Answer 1

Answer:

$\frac{x^2}{8} + \frac{y^2}{4}=1$

Step-by-step explanation:

The directrices in this case are vertical lines, so we have a horizontal ellipse. The equation for that ellipse is:

$\frac{(x - h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1$

The center of the ellipse is (h,k), the diretrix is x = d and the foci are given by (h+c, k) and (h-c, k)

So, comparing the foci, we have that k = 0 and:

$h + c = 2$

$h - c = -2$

Adding these two equations, we have:

$2h = 0$

$h = 0$

$c = 2$

We can find the value of a^2 using the property:

$c / a = a / d$

Using c = 2 and d = 4, we have:

$a^2 = c * d$

$a^2 = 8$

Now, to find b^2, we use the property:

$a^2 = b^2 + c^2$

$8 = b^2 + 4$

$b^2 = 4$

So the equation of the ellipse is:

$\frac{x^2}{8} + \frac{y^2}{4}=1$