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AshLeyx0O · May 28, 2009, 10:37 AM

how do I find the vertex, the focus, and the directrix of the parabola x^2-8x+4y+36=0

Perito · May 28, 2009, 01:09 PM

Originally Posted by ;

the vertex, the focus, and the directrix of the parabola x^2-8x+4y+36=0

You put the equation in a "standard" form. This is taken from:

Parabola - Wikipedia, the free encyclopedia

"In Cartesian coordinates, a parabola with an axis parallel to the y axis with vertex (h,k), focus (h,k + p), and directrix y = k − p, with p being the distance from the vertex to the focus, has the equation with axis parallel to the y-axis

$(x-h)^{2}=4p(y-k)\,$

or, alternatively with axis parallel to the x-axis

$(y-k)^{2}=4p(x-h)\,$

More generally, a parabola is a curve in the Cartesian plane defined by an irreducible equation of the form

$A x^{2} + B xy + C y^{2} + D x + E y + F = 0 \,$

such that $B^{2} = 4 AC \,$ , where all of the coefficients are real, where $A \not= 0$ or $C \not= 0$ , and where more than one solution, defining a pair of points (x, y) on the parabola, exists. That the equation is irreducible means it does not factor as a product of two not necessarily distinct linear equations"

So, for your equation,

$x^2-8x+4y+36=0$

The X is squared, so we'll try to get it into the first form

$x^2-8x + 16= -4y -20$

$(x-4)^2=-4(y-5)$

so h=4, p=-1, k=5

vertex (h,k) = (4,5)
focus (h,k + p) = (4, 4)
directrix y = k − p; y=6
and the axis is parallel to the y-axis

rudolphtaylor · May 31, 2012, 05:24 PM

-x2+y2+4y-16=0