Question

Can anyone tell me the standard form of this parabola equation?

Answer 1

Answer:

$x=4y^2-24y+30$

$(x-4.5)^2+(y-2.5)^2=23$

Step-by-step explanation:

Standard form of a sideways parabola:  $x=ay^2+by+c$

Given equation:

$4y^2-x-24y+30=0$

Add x to both sides:

$4y^2-24y+30=x$

$\implies x=4y^2-24y+30$

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Standard form of circle equation

$(x-a)^2+(y-b)^2=r^2$

(where (a,b) is the center and r is the radius)

Given equation:

$2x^2+2y^2-18x-10y+7=0$

Group like terms:

$2x^2-18x+2y^2-10y+7=0$

Divide by 2:

$x^2-9x+y^2-5y+3.5=0$

Factor by completing the square for each variable:

$(x-4.5)^2-20.25+(y-2.5)^2-6.25+3.5=0$

Rearrange into standard form:

$(x-4.5)^2+(y-2.5)^2=23$

Therefore, the circle has a center at (4.5, 2.5) and a radius of √23