Question

1.Write the equation for an ellipse with vertices (–4, 2), (2, 2), (–1, –2) and (–1, 6).

Answer 1

The general equation of an ellipse:
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} =1$

we have vertices (–4, 2), (2, 2), (–1, –2) and (–1, 6).⇒⇒⇒ red points
by graphing the points ⇒⇒⇒ attached figure 1
the majority axis is the line connecting (–1, –2) and (–1, 6) and has a distance = 8
the minority axis is the line connecting (–4, 2), and (2, 2) and has a distance = 6
(h,k) represents the center of ellipse which is the intersection between axes
∴(h,k) = (-1,2)
and a = 3 , b = 4
∴ the equation of the ellipse is

$\frac{(x+1)^2}{9} + \frac{(y-2)^2}{16} =1$
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A hyperbola with vertices (9, 3) and (5, 3) ⇒⇒⇒⇒ blue points
and with foci (11, 3) and (3, 3). ⇒⇒⇒⇒ red points
by graphing the points ⇒⇒⇒ attached figure 2
so, the hyperbole axis is horizontal
(h,k) represents the center of hyperbola = (7,3) ⇒⇒⇒ green point
a = distance between center and any of vertices = 7 - 5 = 2
c = distance between center and any of foci = 7 - 3 = 4
∵ c² = a² + b²
∴ b² = c² - a² = 16 - 4 = 12
the general equation of the hyperbole :
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =1$
the equation of the hyperbole will be
$\frac{(x-7)^2}{4} - \frac{(y-3)^2}{12} =1$