Question

What is the standard form given the vertex (-3,3) and the focus point (-3,2)

Answer 1

Answer:

$y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{3}{4}$

Step-by-step explanation:

The standard form of a parabola is written as

$y=ax^2+bx+c$

where a, b and c are the coefficients of the second-degree, first degree and zero-degree terms.

The coordinates of the vertex of a parabola is given by:

$x_b = -\frac{b}{2a}$

$y_b=c-\frac{b^2}{4a}$

The coordinates of the focus instead are given by

$x_f=-\frac{b}{2a}$

$y_f=y_b+\frac{1}{4a}$

In this problem, we know the coordinates of the vertex and of the focus point:

Vertex: (-3,3)

Focus point: (-3,2)

So we have:

$x_b=-3=-\frac{b}{2a}$ (1)

$y_b=3=c-\frac{b^2}{4a}$ (2)

$y_f=2=y_b+\frac{1}{4a}$ (3)

From eq.(1) we get

$2a=\frac{b}{3}$ (4)

Substituting into (2),

$3=c-\frac{b^2}{2(b/3)}\\3=c-\frac{3}{2}b\\c=3+\frac{3}{2}b$(5)

Now rewriting eq.(3) as

$2=y_b+\frac{1}{4a}\\2=(c-\frac{3}{2}b)+\frac{1}{4a}$

And substituting (4) and (5) into this, we can find b:

$2=((3+\frac{3}{2}b)-\frac{3}{2}b)+\frac{1}{2(b/3)}\\2=3+\frac{3}{2b}\\-1=\frac{3}{2b}\\b=-\frac{3}{2}$

Then we can find a and c:

$2a=\frac{b}{3}\\a=\frac{b}{6}=\frac{-3/2}{6}=-\frac{1}{4}$

And

$c=3+\frac{3}{2}b=3+\frac{3}{2}(-\frac{3}{2})=3-\frac{9}{4}=\frac{3}{4}$

So the parabola is

$y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{3}{4}$