HSG.GPE.A.3 Math Practice Questions

Practice playeasy

What is the radius of the circle $(x - 1)^{2} + (y - 5)^{2} = 36$ ?

The form $(x - h)^{2} + (y - k)^{2} = r^{2}$ gives $r^{2} = 36$ , so $r = \sqrt{36} = 6$ .

Practice playeasy

For the ellipse $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$ , what is the length of the semi-minor axis?

The smaller denominator is $b^{2} = 9$ , so the semi-minor axis has length $b = \sqrt{9} = 3$ .

Practice playeasy

The hyperbola $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ has a vertex on the positive x-axis. What is the x-coordinate of that vertex?

Set $y = 0$ : then $x^{2} = 16$ , so the vertices are $(\pm 4, 0)$ and the positive x-coordinate is $4$ .

Practice playeasy

Identify the conic: $\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} = 1$ .

Both variables are squared with positive coefficients and different denominators: this is an ellipse.

Practice playeasy

The graph of $y = (x - 3)^{2} + 5$ is a parabola. What is the x-coordinate of its vertex?

Vertex form $y = (x - h)^{2} + k$ has vertex $(h, k)$ , so the vertex is $(3, 5)$ and the x-coordinate is $3$ .

Practice playeasy

The ellipse $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$ crosses the positive x-axis at one point. What is the x-coordinate of that point?

Set $y = 0$ : then $x^{2} = 25$ , and the positive solution is $x = 5$ .

Practice playeasy

The hyperbola $\dfrac{y^{2}}{25} - \dfrac{x^{2}}{4} = 1$ opens up and down. What is the y-coordinate of the vertex on the positive y-axis?

Set $x = 0$ : then $y^{2} = 25$ , so the vertices are $(0, \pm 5)$ and the positive y-coordinate is $5$ .

Practice playeasy

Identify the conic: $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ .

One squared term is subtracted from the other: this is a hyperbola.

HSG.GPE.A.3 math practice. Learn by doing.