Center:
(0, 0); Vertices: the point negative two square root two comma zero
and the point 2 square root two comma zero; Foci: Ordered pair
negative square root 6 comma zero and ordered pair square root 6
comma zero
7. B)
A vertical ellipse is shown on the coordinate plane centered at the
origin with vertices at the point zero comma seven and zero comma
negative seven. The minor axis has endpoints at negative six comma
zero and six comma zero.
8. D)
x squared divided by 16 plus y squared divided by 25 equals 1
Center:
(0, 0); Vertices: the point negative two square root two comma zero
and the point 2 square root two comma zero; Foci: Ordered pair
negative square root 6 comma zero and ordered pair square root 6
comma zero
7. B)
A vertical ellipse is shown on the coordinate plane centered at the
origin with vertices at the point zero comma seven and zero comma
negative seven. The minor axis has endpoints at negative six comma
zero and six comma zero.
8. D)
x squared divided by 16 plus y squared divided by 25 equals 1
Distance formula: the distance, d, between two coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2[/tex]geometric definition of a parabola: set of points on a plane that are the same distance from a focus point and a directrix line.let [tex]d_a[/tex] be the distance between the point [tex](x,y)[/tex] and the focus point at [tex](0,-10)[/tex]. using the distance formula: [tex]\begin{aligned} d_a & = \sqrt{(x-0)^2 + ())^2 \\ & = \sqrt{x^2+ (y+10)^2} \end{aligned}[/tex]let [tex]d_b[/tex] be the distance between the point [tex](x,y)[/tex] and the directrix at [tex]y=10[/tex], which is a horizontal line. when talking about this distance between a point and a horizontal line, we only care about the vertical distance. there is a point on the horizontal line that would share the same x-coordinate. so[tex]\begin{aligned} d_b & = \sqrt{(x-x)^2 + (y-10)^2} \\ & = \sqrt{(y-10)^2}\end{aligned}[/tex]these two distances must be equal: [tex]\begin{aligned} d_a & = d_b \\ \sqrt{x^2+ (y+10)^2} & = \sqrt{(y-10)^2}\end{aligned}[/tex]if we square both sides to get rid of the square roots [tex]\begin{aligned}x^2+ (y+10)^2 & = (y-10)^2 \\ x^2+ y^2+20y+100 & = y^2-20y+100\end{aligned}[/tex]both sides have [tex]y^2[/tex] and 100; if we subtract those two from both sides of the equation: [tex]\begin{aligned}x^2+20y & =-20y \\ x^2 & = -40y \\ y & = -\frac{1}{40}x^2 \end{aligned}[/tex]so [tex]y = -\dfrac{1}{40}x^2[/tex]
The parabola opens to the left, and its vertex is at (0, 0). The focus must have an x-coordinate that is negative. The only viable choice is ...
(-10, 0)
__
The equation is in the form ...
4px = y^2
where p is the distance from the vertex to the focus.
In the given equation, 4p = -40, so p=-10, and the focus is 10 units to the left of the vertex. In this equation, the vertex corresponds to the values of the variables where the squared term is zero: (x, y) = (0, 0).
0
helpful
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⇒ 
where the focus is located at 
. Therefore, the focus of the parabola is 
.
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