Answer to Question #171138 in Analytic Geometry for Jon jay mendoza

Question

Determine the focus and directrix of the parabola with the given equation. Sketch the

graph, indicate the focus, directrix, vertex, and axis of symmetry. (2 items, 20 points)

1. x^2 − 6x − 5y = −34

2. y^2 − 12x + 8y = −40

Answer 1

Solution.

1) $x^2-6x-5y=-34.$

$y=\frac{1}{5}(x-3)^2-5.$

From here vertex of the parabola $(h;k)$ will be $(3;5).$

$x=3$ - axis of symmetry.

Focus $(h;k+\frac{1}{4a}),$ where $a=\frac{1}{5}.$ So, $(3;6.25)$ is the focus of the parabola.

$y=k-\frac{1}{4a}$ is the equation of directrix.

Therefore $y=3.75$ is directrix.

2) $y^2-12x+8y=-40.$

$(y+4)^2=12(x-2).$

From here $(2;-4)$ is the vertex of parabola.

$y=-4$ - axis of symmetry.

Focus $(h+p;k),$ where $p=3.$ So, $(5;-4)$ is the focus.

$x=-1$ is the directrix of the parabola.