High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Grade 12 Pre-Calculus Textbook > Conics [BC/AB]

Hyperbolas

EllipsesPrevious Section
Popular Courses
Find My Course

Hyperbolas

A hyperbola is generated when a plane intersects both nappes of the cone. As a result:
  • a hyperbola has two foci.
  • for every point on the hyperbola, the difference of the distance to each foci is constant.

Hyperbola that Opens Sideways:

The derivation of a hyperbola is similar to an ellipse except we consider the difference of distances
∣PF1∣−∣PF2∣=±2a|PF_1|-|PF_2|=\pm2a to get the equation of a hyperbola:

x2a2−y2b2=1\boxed{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}

such that:
  • the foci are located at (±c,0)(\pm{}c,0)where c2=a2+b2c^2=a^2+b^2
  • the vertices are located at (±a,0)(\pm{}a,0)
  • the asymptotes are y=±(ba)xy=\pm\Big(\frac{b}{a}\Big)x



Wize Tip
The transformed form for a hyperbola is (x−h)2a2−(y−k)2b2=1\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1

where:
  • the asymptotes are y=k±ba(x−h)y=k\pm\frac{b}{a}(x-h)
  • the vertices are (±a+h,0)(\pm{a}+h,0)
  • the foci are at (±c+h,0)(\pm{c}+h,0), where c=a2+b2c=\sqrt{a^2+b^2}

PAGE BREAK

Hyperbola that Opens Up/Down:

y2a2−x2b2=1\boxed{\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1}

such that:
  • the foci are located at (0,±c)(0,\pm{}c)where c2=a2+b2c^2=a^2+b^2
  • the vertices are located at (0,±a)(0,\pm{}a)
  • the asymptotes are y=±(ab)xy=\pm\Big(\frac{a}{b}\Big)x


Wize Tip
The transformed form for a hyperbola is (y−k)2a2−(x−h)2b2=1\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1

where
  • the asymptotes are y=k±ab(x−h)y=k\pm\frac{a}{b}(x-h)
  • the vertices are (0,±a+k)(0,\pm{a}+k)
  • the foci are (0,±c+k)(0,\pm{c}+k), where c=a2+b2c=\sqrt{a^2+b^2}

Example: Hyperbolas

Sketch the hyperbola 9x2−4y2+36x−8y−4=09x^2-4y^2+36x-8y-4=0.

First, complete the square to get it into standard form:

9x2−4y2+36x−8y−4=0(9x2+36x)+(−4y2−8y)=49(x2+4x+4)−4(y2+2y+1)=4+36−49(x+2)2−4(y+1)2=36(x+2)24−(y+1)29=1\begin{array}{rcl} 9x^2-4y^2+36x-8y-4&=&0\\\\ (9x^2+36x)+(-4y^2-8y)&=&4\\\\ 9(x^2+4x+4)-4(y^2+2y+1)&=&4+36-4\\\\ 9(x+2)^2-4(y+1)^2&=&36\\\\ \dfrac{(x+2)^2}{4}-\dfrac{(y+1)^2}{9}&=&1 \end{array}


The asymptotes are at:

y=−1±32(x+2)∴ y1=32x+2y2=−32x−4\begin{array}{rcl} y&=&-1\pm\dfrac{3}{2}(x+2)\\\\ \therefore~y_1&=&\dfrac{3}{2}x+2\\\\ y_2&=&-\dfrac{3}{2}x-4 \end{array}


The center is located at (−2,−1)(-2,-1)

The vertices are located ±2\pm2 units horizontally away from the center: (−4,−1)(-4,-1) and (0,−1)(0,-1).

Therefore,



Practice: Hyperbolas

Find the following information for the hyperbola 9x2−16y2=144.9x^2-16y^2=144.

Practice: Hyperbolas

Find the following information for the hyperbola 9x2−16y2=144.9x^2-16y^2=144.

Practice: Hyperbolas

Find the equation and the foci of the hyperbola with vertices (0,±1)(0,\pm1) and asymptotes y=±2xy=\pm{2x}.

Practice: Hyperbolas

Find the center, foci, vertices, and equations of asymptotes for the hyperbola 9x2−16y2−18x+96y+9=09x^2-16y^2-18x+96y+9=0.
Extra Practice
Conics
Let 25y2+250y−16x2−32x+209=025y^2+250y−16x^2−32x+209=0.
Determine the following:
Hyperbolas
Practice: Hyperbolas
Find the center, foci, vertices, and equations of asymptotes for the hyperbola −16x2+64x+4y2+16y=64-16x^2+64x+4y^2+16y=64.
Hyperbolas
Practice: Hyperbolas
Find the center, foci, vertices, and equations of asymptotes for the hyperbola
4x2+40x−5y2−30y=454x^2+40x-5y^2-30y=45.
Hyperbolas
Practice: Hyperbolas
Find the equation and the foci of the hyperbola with vertices (4,5)(4,5), (−7,5)(-7,5)and asymptotes y=±37xy=\pm{\dfrac{3}{7}x}.
Hyperbolas
Practice: Hyperbolas
Find the equation and the foci of the hyperbola with vertices (−6,3)(-6,3), (4,3)(4,3)and asymptotes y=±73xy=\pm{\dfrac{7}{3}x}.
Hyperbolas
Practice: Hyperbolas
Find the following information for the hyperbola 9x2−16y2=144.9x^2-16y^2=144.
Hyperbolas
Practice: Hyperbolas
Find the following information for the hyperbola 5x2−25y2=625.5x^2-25y^2=625.