Given: #2y^2 - 32x + 20y + 50 = 0#
The equation given is in general form. Since there is no #x^2# term, the equation can't be a circle, ellipse or hyperbola. That leaves only the parabola as the only possibility.
One way is to actually solve for #y# by completing the square and graph the function:
#(2y^2 + 20y) - 32x = -50#
#2(y^2 + 10y) - 32x = -50#
Complete the square by halving the #y# coefficient and then adding #2*# the square of that coefficient to the other side:
#2(y +5)^2 -32x = -50 + 2(5^2)#
#2(y+5)^2 - 32x = 0#
#2(y+5)^2 = 32x#
#(y+5)^2 = 32/2 x = 16x#
square root both sides: #" "y+5 = +- sqrt(16x)#
#y = -5 +- 4sqrt(x)#
graph both pieces: #Y1 = -5 + 4 sqrt(x); " "Y2 = -5 - 4sqrt(x)#
Horizontal axis parabola: #x =a (y - k)^2 + h #; vertex #(h, k); a# is a constant
From above, #" "x = 1/16(y+5)^2; " vertex "(0, -5)#
A second way to find out for sure is to analyze the coefficients in the equation as follows:
The general equation of conics:
#ax^2 + bxy + cy^2 + dx + ey + f = 0#
If the conic is rotated it will have a #b# term. Let's assume in the standard case we don't have any rotations of conics. This means #b = 0#.
Vertical parabola when #b = c = 0, a!= 0, e !=0# in the form: #ax^2 + dx + ey + f = 0#
Horizontal parabola when #a = b = 0, c != 0, d != 0# in the form: #cy^2 + dx + ey + f = 0#
Circle when #b = 0; a !=0; c!=0: " "ax^2 + cy^2 + dx + ey +f = 0#
Ellipse when #b = 0; a " & "c > 0# or #a " & "c < 0, a !=c#
Hyperbola when #b = 0, a " & "c # have opposite signs. The general form is #x^2 - cy^2 + dx + ey = 0#
