First find all the goods on 25x^2 + 16y^2 - 150x - 32y = 159
by completing the square
25(x^2 + 6x + .....) + 16(y^2 - 2y + ....) = 159
25(x^2 + 6x + 9) + 16(y^2 - 2y + 1) = 159 + 225 + 16
25(x+3)^2 + 16(y-1)^2 = 400
divide each term by 400
(x+3)^2 / 16 + (y-1)^2 / 25 = 1
so we have a vertical ellipse, with centre at (-3,1)
vertices at the end of the major axis, those being (-3,5) and (-3,-5)
and at the end of the minor axis at (4,1) and (-4,1)
for the foci:
c^2 = 5^2 - 4^2 = 9
c = ± 3
The two foci would be 3 units above and below the centre(-3,1)
which would be (-3,4) and (-3,-2)
Our parabola must have as a vertex the focus of the ellipse, so one would
be:
y = a(x+3)^2 + 4 , and the other would be y = a(x+3)^2 - 2
the first:
y = a(x+3)^2 + 4
the shape of our parabola is determined by x^2 = 4py ,
(a standard formula) , but we know p = 1
(the difference between the vertex and the focus ) , so x^2 = 4y or y = (1/4)x^2
our parabolas are:
y = (1/4)(x+3)^2 + 4 and y = (1/4)(x+3)^2 - 2
An ellipse has equation 25𝑥^2 + 16𝑦^2 + 150𝑥 − 32𝑦 = 159. Find the standard equations of all parabolas whose vertex is a focus of this ellipse and whose focus is a vertex of this ellipse
1 answer