High School Math : Hyperbolas

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #2 : Pre Calculus

A conic section is represented by the following equation:

\(\displaystyle \small -4x^{2} + 15y^{2} + 343 =0\)

What type of conic section does this equation represent?

Possible Answers:

Ellipse

Circle

Parabola

Hyperbola

Correct answer:

Hyperbola

Explanation:

The simplest way to know what kind of conic section an equation represents is by checking the coefficients in front of each variable. The equation must be in general form while you do this check. Luckily, this equation is already in general form, so it's easy to see. The general equation for a conic section is the following:

\(\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\)

Assuming the term \(\displaystyle Bxy\) is 0 (which it usually is):

  • If A equals C, the equation is a circle.
  • If A and C have the same sign (but are not equal to each other), the equation is an ellipse.
  • If either A or C equals 0, the equation is a parabola.
  • If A and C are different signs (i.e. one is negative and one is positive), the equation is a hyperbola.

Example Question #1 : Pre Calculus

A conic section is represented by the following equation:

 

\(\displaystyle \frac{(x-3)^{2}}{4}-\frac{(y+2)^{2}}{9}=1\)

Which of the following best describes this equation?

Possible Answers:

horizontal hyperbola with center of \(\displaystyle (3,-2)\) and asymptotes with slopes of \(\displaystyle \frac{3}{2}\) and \(\displaystyle -\frac{3}{2}\)

 

vertical hyperbola with center \(\displaystyle (3, -2)\) and asymptotes with slopes of \(\displaystyle \frac{2}{3}\) and \(\displaystyle -\frac{3}{2}\)

vertical parabola with vertex \(\displaystyle (-3, 2)\) and a vertical stretch factor of \(\displaystyle 4\)

vertical ellipse with center \(\displaystyle (-3,2)\) and a major axis length of \(\displaystyle 6\)

horizontal hyperbola with center \(\displaystyle (3,-2)\) and asymptotes with slopes of \(\displaystyle \frac{2}{3}\) and \(\displaystyle -\frac{2}{3}\)

Correct answer:

horizontal hyperbola with center of \(\displaystyle (3,-2)\) and asymptotes with slopes of \(\displaystyle \frac{3}{2}\) and \(\displaystyle -\frac{3}{2}\)

 

Explanation:

First, we need to make sure the conic section equation is in a form we recognize. Luckily, this equation is already in standard form:

\(\displaystyle \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)

The first step is to determine the type of conic section this equation represents. Because there are two squared variables (\(\displaystyle x^{2}\) and \(\displaystyle y^{2}\)), this equation cannot be a parabola.  Because the coefficients in front of the squared variables are different signs (i.e. one is negative and the other is positive), this equation must be a hyperbola, not an ellipse.

In a hyperbola, the squared term with a positive coefficient represents the direction in which the hyperbola opens. In other words, if the \(\displaystyle x^{2}\) term is positive, the hyperbola opens horizontally. If the \(\displaystyle y^{2}\) term is positive, the hyperbola opens vertically. Therefore, this is a horizontal hyperbola.

The center is always found at \(\displaystyle (h, k)\), which in this case is \(\displaystyle (3, -2)\).

That leaves only the asymptotes. For a hyperbola, the slopes of the asymptotes can be found by dividing \(\displaystyle b\) by \(\displaystyle a\) (remember to always put the vertical value, \(\displaystyle b\), above the horizontal value, \(\displaystyle a\)). Remember that these slopes always come in pairs, with one being positive and the other being negative.

In this case, \(\displaystyle b\) is 3 and \(\displaystyle a\) is 2, so we get slopes of \(\displaystyle \frac{3}{2}\) and \(\displaystyle -\frac{3}{2}\).

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