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Classifying Conic Sections

Use the discriminant to classify ellipses, parabolas, and hyperbolas

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Classifying Conic Sections

You and your friends are playing Name the Conic Section. Your friend pulls a card with the equation \begin{align*}x^2 + 3xy = -5y^2 -10\end{align*} written on it. What type of conic section is represented by the equation?

Classifying Conic Sections

Another way to classify a conic section when it is in the general form is to use the discriminant, like from the Quadratic Formula. The discriminant is what is underneath the radical, \begin{align*}b^2-4ac\end{align*}, and we can use this to determine if the conic is a parabola, circle, ellipse, or hyperbola. If the general form of the equation is \begin{align*}Ax^2+Bxy+Cy^2+Dx+Ey+F=0\end{align*}, where \begin{align*}B=0\end{align*}, then the discriminant will be \begin{align*}B^2-4AC\end{align*}.

Use the table below:

\begin{align*}B^2-4AC=0\end{align*} and \begin{align*}A=0\end{align*} or \begin{align*}C=0\end{align*} Parabola
\begin{align*}B^2-4AC<0\end{align*} and \begin{align*}A=C\end{align*} Circle
\begin{align*}B^2-4AC<0\end{align*} and \begin{align*}A \ne C\end{align*} Ellipse
\begin{align*}B^2-4AC>0\end{align*} Hyperbola

Let's use the discriminant to determine the type of conic section for the following equations.

  1. \begin{align*}x^2-4y^2+5x-8y+16=0\end{align*}

\begin{align*}A=1\end{align*}, \begin{align*}B=0\end{align*}, and \begin{align*}C=-4\end{align*}

\begin{align*}0^2-4(1)(-4)=16\end{align*} This is a hyperbola.

  1. \begin{align*}3x^2+3y^2-9x-12y-20=0\end{align*}

\begin{align*}A=3\end{align*}, \begin{align*}B=0\end{align*}, \begin{align*}C=3\end{align*}

\begin{align*}0^2-4(3)(3)=-36\end{align*} Because \begin{align*}A=C\end{align*} and the discriminant is less than zero, this is a circle.

Finally, let's use the discriminant to determine the type of conic. Then, we'll change the equation into standard form to verify our answer. We'll also find the center (or vertex, if it is a parabola). \begin{align*}x^2+y^2-6x+14y-86=0\end{align*}

 \begin{align*}A=1, \ B=0, \ C=1\end{align*} This is a circle.

\begin{align*}(x^2-6x+9)+(y^2+14y+49)&=86+49+9 \\ (x-3)^2+(y+7)^2 &=144 \end{align*}

The center is \begin{align*}(3, \ -7)\end{align*}.

Examples

Example 1

Earlier, you were asked to determine the type of conic section represented by the equation \begin{align*}x^2 + 3xy = -5y^2 -10\end{align*}

First we need to rewrite the equation is standard form.

\begin{align*}x^2 + 3xy = -5y^2 -10 x^2 + 3xy + 5y^2 + 10 = 0\end{align*}

Now we can use the discriminant to find the type of conic section represented by the equation.

\begin{align*}A=1, \ B=3, \ C=5\end{align*}

\begin{align*}3^2-4(1)(5)=-11\end{align*} Because \begin{align*}A \ne C\end{align*} and the discriminant is less than zero, this equation represents an ellipse.

For Examples 2 & 3, use the discriminant to determine the type of conic.

Example 2

\begin{align*}2x^2+5y^2-8x+25y+115=0\end{align*}

\begin{align*}0^2-4(2)(5)=-40\end{align*}, this is an ellipse.

Example 3

\begin{align*}5y^2-9x-10y-14=0\end{align*}

\begin{align*}0^2-4(0)(5)=0\end{align*}, this is a parabola.

Example 4

Use the discriminant to determine the type of conic. Then, change the equation into standard form to verify your answer. Find the center or vertex, if it is a parabola.

\begin{align*}-4x^2+3y^2-8x+24y+32=0\end{align*}

\begin{align*}0^2-4(-4)(3)=48\end{align*}, this is a hyperbola. Changing it to standard form, we have:

\begin{align*}(-4x^2-8x)+(3y^2+24y) &=-32 \\ -4(x^2+2x+1)+3(y^2+8y+16) &=-32+48-4 \\ -4(x+1)^2+3(y+4)^2 &=12 \\ \frac{-(x+1)^2}{3}+\frac{(y+4)^2}{4} &=1\end{align*}

Usually, we write the negative term second, so the equation is \begin{align*}\frac{(y+4)^2}{4} - \frac{(x+1)^2}{3}=1\end{align*}. The center is \begin{align*}(-1, -4)\end{align*}.

Review

Use the discriminant to determine the type of conic each equation represents.

  1. \begin{align*}2x^2+2y^2+16x-8y+25=0\end{align*}
  2. \begin{align*}x^2-y^2-2x+5y-12=0\end{align*}
  3. \begin{align*}6x^2+y^2-12x+7y+35=0\end{align*}
  4. \begin{align*}3x^2-15x+9y-18=0\end{align*}
  5. \begin{align*}10y^2+6x-40y+253=0\end{align*}
  6. \begin{align*}4x^2+4y^2+32x+48y+465=0\end{align*}

Match the equation with the correct graph.

  1. \begin{align*}x^2+10x+4y+41=0\end{align*}
  2. \begin{align*}4y^2+x+56y+188=0\end{align*}
  3. \begin{align*}x^2+y^2+10x-14y+65=0\end{align*}
  4. \begin{align*}25x^2+y^2-200x-10y+400=0\end{align*}

Use the discriminant to determine the type of conic. Then, change the equation into standard form to verify your answer. Find the center or vertex, if it is a parabola.

  1. \begin{align*}x^2-12x+6y+66=0\end{align*}
  2. \begin{align*}x^2+y^2+2x+2y-2=0\end{align*}
  3. \begin{align*}x^2-y^2-10x-10y-10=0\end{align*}
  4. \begin{align*}y^2-10x+8y+46=0\end{align*}
  5. Find the Area of an Ellipse Graph \begin{align*}x^2+y^2=36\end{align*} and find its area.
    1. Then, graph \begin{align*}\frac{x^2}{36} + \frac{y^2}{25}=1\end{align*} and \begin{align*}\frac{x^2}{25} + \frac{y^2}{36}=1\end{align*} on the same axes.
    2. Do these ellipses have the same area? Why or why not?
    3. If the equation of the area of a circle is \begin{align*}A=\pi r^2\end{align*}, what do you think the area of an ellipse is? Use \begin{align*}a\end{align*} and \begin{align*}b\end{align*} as in the standard form, \begin{align*}\frac{x^2}{a^2} + \frac{y^2}{b^2}=1\end{align*}.
    4. Find the areas of the ellipses from part a. Are the areas more or less than the area of the circle? Why or why not?

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.11.

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