1) An ellipse has its center at the origin, its foci on the y-axis, and its major axis is three times as long as its minor axis. Given that the ellipse passes through the point (-4,0), find its equation.
2)Find the foci of the ellipse whose major axis has endpoints (0,0) and (13,0) and whose minor axis has length 12.
3) A circle centered at the origin intersects the ellipse y^2/16+x^2/9=1 at the four vertices of a square. Find the area of that square.
Basic fromula for an ellipse x^2/b^2 + y^2/a^2 =1 b = minor axis (given as 4 because ellipse is centered at origin and point 4,0 is on it ...s o b = 4 & b^2 = 16))
a is 3 times b (given) then a = 12 (a^2 =144)
so x^2/16 + y^2/144 =1
I had to do this one graphically to figure out how to deduce this.....
Since the circle is on the origin too, it is of the form
x^2 +y^2 = r^2
now for a square to be on this circle (centered on the origin, along with the ellipse), the x's and the y's absolute value in all of the quadrants must be equal (think about it)
so the point that the line uintersects the ellipse will be x,y but x=y so x,x
This point has to be on the ellipse, so let's substitute this in to the ellipse equation and solve for x (and also y)
x^2/16 + x^2/9 = 1
9x^2/144 + 16x^2/144 = 1
9x^2 + 16x^2 = 144
25 x^2 = 144
x^2 = 144/25 x = 12/5 = y = 2.4
Now since we have the x and y coordinates of a point on the circle we can calculate r^2 = 2.4^2 + 2.4^2 = 2(2.4^2) = 11.52
Then the circle becomes x^2 + y^2 = 11.52
So side of square is 2 x 2.4 = 4.8 area = 4.8 x 4.8 = 23.04 sq units
(there probably is a better way!)