rewrite 2y^2-x-20y+36=0 in vertex form. Determine the type of conic the equation represents?

Question

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Answer 1

Given equation 2y^2-x-20y+36 = 0

Add x to each side.

2y^2-x-20y+x+36 = 0+x

x = 2y^2-20y+36

Compare it to horizontal form of parabola is x = ay^2+by+c

a = 2, b = 20, c = 36

Vertex form is x = a(y-k)^2+h

x = 2y^2-20y+36

to complete square add 50 to each side.

x+50 = 2y^2-20y+36+50

x+50 = 2y^2-20y+50+36

x+50 = 2(y^2-10y+25)+36

Subtract 50 from each side.

x+50-50 = 2(y+5)^2+36-50

x = 2(y+5)^2-14

x = 2(y-(-5))^2+(-14)

Vertex is (h,k) = (-5,-14)

Vertex form is x = 2(y-(-5))^2+(-14).

Answer 2

The equation 2y² - x - 20y + 36 = 0

2y² - 20y + 36 = x

The equation represents horizontal parabola.

General form of horizontal parabola x = ay² + by + c.

a = 2, b = -20 , c = 36.

Vertex form of horizontal parabola is x = a(y - k)² + h.

Where (h ,k) = vertex.

x = 2y² - 20y + 36

Here y ² coefficient is 2 , for perfect square make y ² coefficient 1 by dividing each side by  2.

x/2 = y²  - 10y + 18

To change the expression into a perfect square  add  (half the y coefficient)²to each side of the expression.

Here y coefficient = - 10. so, (half the y coefficient)² = (-10/2)²= 25

x/2 + 25 = y² - 10y + 18 + 25

x/2  = y² - 10y + 25 + 18 - 25

x/2 = (y - 5)² - 7

x = 2(y - 5)² -  2(7)

x = 2(y - 5)² - 14

x = 2(y - 5)² - 14

Vertex form of 2y² - x - 20y + 36 = 0 is x = 2(y - 5)² - 14.