Graphs for Precalculus:: Conics, Polar and Parametric Curves Flashcards (37 Cards) — Study Online

Curriculum Overview

Learn conic sections, parametric and polar curves by reading their graphs. 37 flashcards, each with a real graph on the front: circles and ellipses (center, vertices, foci, eccentricity), parabolas (focus and directrix), hyperbolas (vertices, foci, asymptotes), identifying a conic, parametric curves (eliminating the parameter, direction, a ball in flight, the cycloid, a loop), and polar curves (circles, cardioid, limacon, rose, spiral, lemniscate, conics in polar form). Every answer is checked and explained. For precalculus, AP Calculus BC and Calculus 2.

Topics & Key Concepts

conics circle ellipse parabola hyperbola foci directrix asymptotes eccentricity parametric equations eliminate the parameter cycloid polar coordinates polar graphs cardioid limacon rose curve spiral lemniscate precalculus Calculus 2 AP Calculus BC

Sample Flashcard Questions & Answers

Showing 8 of 37 cards

Question #1 Active Recall

The graph of a circle centered at the origin is shown. Write its equation.

Answer & Explanation:

\(x^2+y^2=16\).

The circle crosses the axes at \((\pm 4, 0)\) and \((0, \pm 4)\), so the radius is \(r=4\) and \(x^2+y^2=r^2=16\).

Question #2 Active Recall

The graph of a circle is shown. Its center is at \((2,-1)\) and it passes through \((5,-1)\). Write its equation.

Answer & Explanation:

\((x-2)^2+(y+1)^2=9\).

The radius is the distance from the center \((2,-1)\) to \((5,-1)\): \(r=3\). The standard form is \((x-h)^2+(y-k)^2=r^2\) with \((h,k)=(2,-1)\).

Question #3 Active Recall

The graph of \(x^2+y^2-4x+6y-12=0\) is shown. Find the center and the radius.

Answer & Explanation:

Center \((2,-3)\), radius \(5\).

Complete the square: \((x^2-4x+4)+(y^2+6y+9)=12+4+9\), so \((x-2)^2+(y+3)^2=25\). The circle crosses the x-axis at \((-2,0)\) and \((6,0)\) (set \(y=0\): \((x-2)^2=16\)), and its lowest point is \((2,-8)\), five units below the center.

Question #4 Active Recall

The graph of an ellipse centered at the origin is shown. It crosses the x-axis at \((\pm5,0)\) and the y-axis at \((0,\pm3)\). Write its equation and find its foci.

Answer & Explanation:

Equation: \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\). Foci: \((\pm4, 0)\).

Here \(a=5\) (the longer half-axis) and \(b=3\). The foci lie on the major axis at distance \(c=\sqrt{a^2-b^2}=\sqrt{25-9}=4\) from the center. The eccentricity is \(e=\dfrac{c}{a}=\dfrac45\).

Question #5 Active Recall

The graph of an ellipse centered at the origin is shown. It crosses the x-axis at \((\pm2,0)\) and the y-axis at \((0,\pm4)\). Write its equation and find its foci.

Answer & Explanation:

Equation: \(\dfrac{x^2}{4}+\dfrac{y^2}{16}=1\). Foci: \((0,\pm2\sqrt3)\), about \((0,\pm3.46)\).

The major axis is vertical (the longer half-axis, 4, is along the y-axis): \(a=4\), \(b=2\). Then \(c=\sqrt{16-4}=\sqrt{12}=2\sqrt3\), and the foci are on the y-axis. The eccentricity is \(\dfrac{c}{a}=\dfrac{\sqrt3}{2}\approx 0.87\).

Question #6 Active Recall

The graph of \(\dfrac{(x-1)^2}{9}+\dfrac{(y+2)^2}{4}=1\) is shown. Find the center, the vertices and the foci.

Answer & Explanation:

Center \((1,-2)\). Vertices \((-2,-2)\) and \((4,-2)\) (the ends of the major axis, \(a=3\)). Foci \((1\pm\sqrt5,-2)\), about \((-1.24,-2)\) and \((3.24,-2)\).

The denominators give \(a^2=9\) (under \(x\), so the major axis is horizontal) and \(b^2=4\); the co-vertices are \((1,0)\) and \((1,-4)\). \(c=\sqrt{9-4}=\sqrt5\).

Question #7 Active Recall

The graph shows two ellipses: \(\dfrac{x^2}{25}+\dfrac{y^2}{24}=1\) (blue) and \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\) (red). Find the eccentricity of each. Which is closer to a circle?

Answer & Explanation:

Blue: \(c=\sqrt{25-24}=1\), \(e=\dfrac15=0.2\). Red: \(c=\sqrt{25-9}=4\), \(e=\dfrac45=0.8\).

The **blue** ellipse is closer to a circle: its eccentricity is closer to 0 (a circle has \(e=0\)). The red ellipse is much more stretched out; as \(e\) approaches 1 an ellipse becomes flatter.

Question #8 Active Recall

The graph of a parabola with its vertex at the origin is shown. It passes through \((2,4)\) and \((2,-4)\). Write its equation and find its focus and directrix.

Answer & Explanation:

Equation: \(y^2=8x\). Focus: \((2,0)\). Directrix: \(x=-2\).

The parabola opens to the right, so \(y^2=4px\). Using the point \((2,4)\): \(16=4p\cdot 2\), so \(4p=8\) and \(p=2\). The focus is \((p,0)=(2,0)\) and the directrix is \(x=-p=-2\). Every point of the parabola is the same distance from the focus and from the directrix.

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