y^2+4y−36x−176=0 is the equation of a parabola. Write the standard form of that parabola.

Question

1).y^2+4y−36x−176=0 is the equation of a parabola. Write the standard form of that parabola.

= _______________________

2).x^2+14x+28y+161=0 is the equation of a parabola. Write the standard form of that parabola.

=_________________________

Answer 1

y^2 + 4y -36x -176 = 0

complete the square, add 4 to both sides

(y^2 +4y +4) = 36x +176 + 4

(y+4)^2 = 36x + 180

36x = (y+4)^2 - 180

x = (1/36)(y+4)^2 - 5 is a rightward opening parabola with vertex = (-5, -4)

that's a standard vertex form: x=a(y-k)^2 +h where (h,k) is the vertex

axis of symmetry is y=-4

when a>0, the parabola opens to the right

or

you could write solving for y, which is another standard form

y = + 6sqr(x +5) - 4

another standard form is x=ay^2 + by + c, it's what you want when you use the quadratic formula to solve for x

x = (1/36)y^2 + (1/9)y - 44/9

2nd problem:

solve for x

x^2 + 14x + 28y + 161 = 0

complete the square

(x+7)^2 = -28y -161 + 49 = -28y - 112

x+7 = + or - 2sqr(-7y-28)

x = + 2sqr(-7(y+4)) -7 which is a downward opening parabola with vertex (-7,-4)

or solve for y which is more standard for a vertical parabola

-28y = (x+7)^2 + 112

y = (-1/28)(x+7)^2 - 4

which is in standard vertex form

or if you want the y=ax^2 + bx + c standard form so you can use the quadratic formula

then

y= (-1/28)x^2 - (1/2)x - 23/4

y=a(x+h)^2 + k where (h,k) is the vertex. when a<0, the parabola is downward opening

axis of symmetry is x= h = -7