Notes
Conics
- Circle: $(x - h)^2 + (y - k)^2 = r^2$. Complete the square to find the center and radius.
- Parabola: $x^2 = 4py$ has focus $(0, p)$ and directrix $y = -p$; $y^2 = 4px$ has focus $(p, 0)$ and directrix $x = -p$.
- Ellipse: $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$ has $c^2 = a^2 - b^2$.
- Hyperbola: $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has $c^2 = a^2 + b^2$ and asymptotes $\displaystyle y = \pm\frac{b}{a}x$.
- Classify by the $x^2$ and $y^2$ coefficients: one missing is a parabola, equal is a circle, same sign is an ellipse, opposite signs is a hyperbola.
Parametric equations
- Substitute values of $t$ to find points. Eliminate $t$ by substitution, or by using $\cos^2 t + \sin^2 t = 1$ for circles and ellipses.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$(x + 4)^2 + (y - 1)^2 = 8 + 16 + 1 = 25$.
$4p = -8$ gives $p = -2$, and the focus is $(p, 0)$.
$\displaystyle c = \sqrt{25 - 9} = 4$.
The slopes are $\displaystyle \pm\frac{b}{a} = \pm\frac{3}{4}$.
$t = x - 1$, so $y = (x - 1)^2$.
Dividing by 4 gives $x^2 + y^2 = 5$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Consider the conic $9x^2 + 16y^2 = 144$.
(a) Write it in standard form and identify the conic, its vertices and its co-vertices.
Show the answer and scoring guide
Divide by 144: $\displaystyle \frac{x^2}{16} + \frac{y^2}{9} = 1$. It is an ellipse with $a = 4$, $b = 3$. The vertices are $(\pm 4, 0)$ and the co-vertices are $(0, \pm 3)$.
Check off the points you earned:
(b) Find the foci and the eccentricity.
Show the answer and scoring guide
$c^2 = 16 - 9 = 7$, so the foci are $\displaystyle (\pm\sqrt{7}, 0)$. Eccentricity $\displaystyle e = \frac{c}{a} = \frac{\sqrt{7}}{4}$.
Check off the points you earned:
(c) Write parametric equations for the ellipse and find the point where $t = \frac{\pi}{3}$.
Show the answer and scoring guide
$x = 4\cos t$, $y = 3\sin t$. At $\displaystyle t = \frac{\pi}{3}$: $\displaystyle x = 4 \cdot \frac{1}{2} = 2$ and $\displaystyle y = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$, so the point is $\displaystyle \left(2, \frac{3\sqrt{3}}{2}\right)$.
Check off the points you earned: