Conic Sections — SAT Math practice questions

Question 1

For the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$, the asymptotes have equations:

Accepted Answer

$a = 3$, $b = 4$, so asymptotes: $y = \pm \dfrac{4}{3}x$.

Answer

$y = \pm x$

Answer

$y = \pm 3x/4$

Answer

$y = \pm 4x/3$

Answer

$y = \pm 5x/3$

Question 2

The parabola $y^2 = 8x$ has focus at:

Accepted Answer

$4p = 8 \Rightarrow p = 2$. Focus: $(2, 0)$.

Answer

$(0, 2)$

Answer

$(2, 2)$

Answer

$(2, 0)$

Answer

$(8, 0)$

Question 3

A circle centered at the origin passes through $(3, 4)$. Its equation is:

Accepted Answer

$r = \sqrt{9 + 16} = 5$, so $x^2 + y^2 = 25$.

Answer

$x^2 + y^2 = 5$

Answer

$x^2 + y^2 = 7$

Answer

$x^2 + y^2 = 25$

Answer

$(x - 3)^2 + (y - 4)^2 = 25$

Question 4

For the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$, the lengths of the semi-axes are:

Accepted Answer

$a^2 = 16 \Rightarrow a = 4$; $b^2 = 9 \Rightarrow b = 3$.

Answer

$a = 16$, $b = 9$

Answer

$a = 8$, $b = 6$

Answer

$a = 4$, $b = 3$

Answer

$a = 2$, $b = \sqrt{3}$

Question 5

Find the radius of the circle $x^2 + y^2 - 4x + 6y + 4 = 0$.

Accepted Answer

$(x^2 - 4x) + (y^2 + 6y) = -4 \Rightarrow (x - 2)^2 - 4 + (y + 3)^2 - 9 = -4 \Rightarrow (x - 2)^2 + (y + 3)^2 = 9$. Radius $= 3$.

Answer

$1$

Answer

$2$

Answer

$3$

Answer

$4$

Question 6

The ellipse $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$ has semi-major axis of length:

Accepted Answer

Here $a^{2} = 25$ and $b^{2} = 9$, so $a = 5$ and $b = 3$.

The semi-major axis is $\max(a, b) = 5$.

Answer

$3$

Answer

$4$

Answer

$5$

Answer

$9$

Question 7

The hyperbola $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ has vertices at:

Accepted Answer

$a^{2} = 16$, so $a = 4$. The vertices are $(\pm a, 0) = (\pm 4, 0)$.

Answer

$(\pm 4,\ 0)$

Answer

$(0,\ \pm 4)$

Answer

$(\pm 3,\ 0)$

Answer

$(\pm 5,\ 0)$

Question 8

The vertex of the parabola $y = x^{2} - 4x + 7$ is at:

Accepted Answer

For $y = ax^{2} + bx + c$, the vertex is at $x = -\dfrac{b}{2a}$.

$x = -\dfrac{-4}{2(1)} = 2$.

$y = (2)^{2} - 4(2) + 7 = 4 - 8 + 7 = 3$.

Vertex: $(2, 3)$.

Answer

$(4, 7)$

Answer

$(2, 3)$

Answer

$(-2, 3)$

Answer

$(2, -3)$

Question 9

The radius of the circle $x^2 + y^2 = 25$ is:

Accepted Answer

The standard form $x^2 + y^2 = r^2$ gives $r = \sqrt{25} = 5$.

Answer

$5/2$

Answer

$\sqrt{5}$

Answer

$5$

Answer

$25$

Question 10

The center of the circle $(x - 2)^2 + (y + 3)^2 = 16$ is:

Accepted Answer

Reading off: $h = 2$, $k = -3$. Center: $(2, -3)$.

Answer

$(-2, 3)$

Answer

$(2, 3)$

Answer

$(2, -3)$

Answer

$(-2, -3)$

Conic Sections practice questions — SAT Math

Problems & worked solutions

Problem 1 — Conic Sections

Problem 2 — Conic Sections

Problem 3 — Conic Sections

Problem 4 — Conic Sections

Problem 5 — Conic Sections

Problem 6 — Conic Sections

Problem 7 — Conic Sections

Problem 8 — Conic Sections

Problem 9 — Conic Sections

Problem 10 — Conic Sections