Problems Involving Conic Sections

Question

Find the standard equation of the parabola opening to the left whose axis contains the major axis of the ellipse 4x² +9y² =144, whose focus is the center of the ellipse, and which passes through the co-vertices of this ellipse.

Answer 1

4x^2 + 9y^2 = 144

4x^2/144 + 9y^2/144 = 1

x^2/36 + y^2/16 = 1

x^2/6^2 + y^2/4^2 = 1

major ellipse axis has co-vertices (-6, 0) and (6, 0), which is the x axis

center of the ellipse is the origin (0,0) = the focus of the parabola

the parabola passes through the two points (-6,0) and (6,0)

the parabola opens to the left, so its standard equation is x = a(y-k)^2 + h where (h,k) is the vertex of the parabola, (h,k) is to the right of the origin with y coordinate = 0

distance from (0,0) to (6,0) is the same distance as from (0,0) to the directrix.

distance from the origin to the directrix = 6. distance from the origin to the vertex is 6/2 = 3

the vertex is (0,3)

the parabola is x= a(y-0)^2 + 3 = ay^2 +3

x = ay^2 +3. plug in either covertex to calculate "a"

0 = a(6^2) + 3 = 36a +3

a = -3/36 = -1/12

the parabola is x = (-1/12)y^2 + 3

or

12x =-y^2 + 36

or

12x +y^2 = 36