The Lumenary · Grades 9–12

Conic sections

Name: ______________________

Date: ______________________

You will learn to

Describe each conic as a section of a double cone and as a locus defined by distances to foci, or to a focus and a directrix
Convert between geometric descriptions and standard equations, completing the square where necessary
Prove and use the reflective properties of the parabola and of the ellipse
Classify second-degree equations by eccentricity and by the sign of $B^2 - 4AC$ , including degenerate cases

1.
The parabola $x^2 = 12y$ has its focus at $(0, p)$ . What is $p$ ?
2.
For every point $P$ on the ellipse $\dfrac{x^2}{49} + \dfrac{y^2}{24} = 1$ , what is $PF_1 + PF_2$ , the sum of the distances to the two foci?
3.
The ellipse $16x^2 + 25y^2 = 400$ has foci at $(\pm c, 0)$ . Find $c$ .
4.
The hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
5.
The equation $2x^2 + 3xy + y^2 + x - 5 = 0$ describes a non-degenerate conic. Which one?
- A.A hyperbola
- B.An ellipse
- C.A parabola
- D.A circle
6.
The equation $x^2 + y^2 - 6x + 8y = 0$ describes a circle. What is its radius?
7.
Which of these equations describe something other than an ellipse, parabola or hyperbola? (Choose all that apply.)
- A. $x^2 + 2y^2 + 3 = 0$
- B. $x^2 - 4y^2 = 0$
- C. $x^2 - 4y^2 = 4$
- D. $(x + y)^2 = 0$
- E. $y = x^2 + x$
8.
In the proof of the reflective property of the parabola, $D$ is the foot of the perpendicular from $P$ to the directrix, and the tangent $t$ at $P$ is the perpendicular bisector of $FD$ . Why does this show that the reflected ray passes through $F$ ?
- A.Reflection in $t$ fixes $P$ and swaps $D$ and $F$ , so it turns the straight-on path from $P$ towards $D$ into the path from $P$ towards $F$
- B.Because $PF = PD$ , the ray has the same length before and after reflection
- C.Because $t$ is parallel to the axis of the parabola
- D.Because $F$ lies on the tangent line $t$
9.
Which Lean statement correctly says that $(x, y)$ is equidistant from the focus $(0, p)$ and the directrix $y = -p$ exactly when $x^2 = 4py$ ?
- A.√(x ^ 2 + (y - p) ^ 2) = |y + p| ↔ x ^ 2 = 4 * p * y
- B.√(x ^ 2 + (y - p) ^ 2) = y + p ↔ x ^ 2 = 4 * p * y
- C.√(x ^ 2 + (y - p) ^ 2) = |y + p| → x ^ 2 = 4 * p * y
- D.√(x ^ 2 + (y + p) ^ 2) = |y - p| ↔ x ^ 2 = 4 * p * y
10.
Write ellipse_foci(a, b) returning the two foci of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ (with $a, b > 0$ ) as a sorted list of two (x, y) tuples of floats. Careful: the major axis is horizontal only when $a \ge b$ !
Computational — Practice

Answer key — Conic sections

1.
Comparing $x^2 = 12y$ with $x^2 = 4py$ gives $4p = 12$ , so $p = 3$ : the focus is $(0, 3)$ and the directrix is $y = -3$ . Check with the point $(6, 3)$ : its distance to the focus is $6$ and to the directrix is $3 + 3 = 6$ .
2.
$a^2 = 49$ , so $a = 7$ and $PF_1 + PF_2 = 2a = 14$ . Check at the vertex $(7, 0)$ : the foci are $(\pm 5, 0)$ (since $c^2 = 49 - 24 = 25$ ), and the distances are $2$ and $12$ .
3.
Dividing by $400$ : $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ , so $a = 5$ , $b = 4$ and $c = \sqrt{25 - 16} = 3$ . The foci are $(\pm 3, 0)$ and the eccentricity is $\tfrac35$ .
4.
$c^2 = a^2 + b^2 = 9 + 16 = 25$ , so $c = 5$ . The eccentricity is $\tfrac53 > 1$ , and the asymptotes are $y = \pm\tfrac43 x$ .
5.
$A = 2$ , $B = 3$ , $C = 1$ , so $B^2 - 4AC = 9 - 8 = 1 > 0$ : a hyperbola. (Indeed the quadratic part factorises as $(2x + y)(x + y)$ , which is zero along two directions — the directions of the asymptotes.)
6.
$(x - 3)^2 - 9 + (y + 4)^2 - 16 = 0$ , so $(x - 3)^2 + (y + 4)^2 = 25$ : centre $(3, -4)$ and radius $5$ . (The circle passes through the origin, as the missing constant term shows.)
7.
$x^2 + 2y^2 + 3 = 0$ is empty, $x^2 - 4y^2 = 0$ is a pair of crossing lines and $(x + y)^2 = 0$ is a single line: all degenerate. $x^2 - 4y^2 = 4$ is a hyperbola and $y = x^2 + x$ is a parabola.
8.
Without the mirror the ray would continue from $P$ towards $D$ . Reflecting in the tangent $t$ , which is the perpendicular bisector of $FD$ , fixes $P$ and swaps $D$ with $F$ . So the reflected ray is the ray from $P$ towards $F$ .
9.
The distance to the focus (0, p) is √(x ^ 2 + (y - p) ^ 2), the distance to the line y = −p is |y + p|, and 'exactly when' is ↔. Dropping the absolute value makes the statement false for negative p; using → states only half of the theorem.
10.
```
import math

def ellipse_foci(a, b):
    c = math.sqrt(abs(a**2 - b**2))
    if a >= b:
        return sorted([(-c, 0.0), (c, 0.0)])
    return sorted([(0.0, -c), (0.0, c)])
```
The foci lie on the major axis, at distance $c = \sqrt{(\text{major})^2 - (\text{minor})^2}$ from the centre. The starter assumes $a \ge b$ ; for $a = 3$ , $b = 5$ it tries to take the square root of a negative number. For a circle ( $a = b$ ) both foci are at the centre.