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Notes
Practice Problems
\(\textbf{1)}\) Graph \( \displaystyle \frac{x^2}{4}+\frac{y^2}{25}=1 \) 
see video for details
see video for details
\(\textbf{2)}\) Graph \( \displaystyle \frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1 \) 
see video for details
see video for details
\(\textbf{3)}\) Find the center of the hyperbola \(\displaystyle\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1\). The center is \((3,-2) \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The center is } (h,k)\)
\(\,\,\,\,\,\,\,\text{The center is } (3,-2)\)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The center is } (h,k)\)
\(\,\,\,\,\,\,\,\text{The center is } (3,-2)\)
\(\textbf{4)}\) Find the vertices of the hyperbola \(\displaystyle\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1\). The vertices are \((-1,-2) \) and \( (7,-2) \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The vertices are } (h\pm a,k)\)
\(\,\,\,\,\,\,\,\text{The vertices are } (3\pm 4,-2)\)
\(\,\,\,\,\,\,\,\text{The vertices are } (-1,-2) \text{ and } (7,-2)\)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The vertices are } (h\pm a,k)\)
\(\,\,\,\,\,\,\,\text{The vertices are } (3\pm 4,-2)\)
\(\,\,\,\,\,\,\,\text{The vertices are } (-1,-2) \text{ and } (7,-2)\)
\(\textbf{5)}\) Find the Foci of the hyperbola \(\displaystyle\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1\).
The foci are \((3+2\sqrt{5},-2) \) and \( (3-2\sqrt{5},-2) \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The foci are } (h\pm c,k)\)
\(\,\,\,\,\,\,\,\text{To solve for c use } c^2=a^2+b^2\)
\(\,\,\,\,\,\,\, c^2=16+4\)
\(\,\,\,\,\,\,\, c^2=20\)
\(\,\,\,\,\,\,\, c=\sqrt{20}\)
\(\,\,\,\,\,\,\, c=2\sqrt{5}\)
\(\,\,\,\,\,\,\,\text{The foci are } (3\pm 2\sqrt{5},-2)\)
The foci are \((3+2\sqrt{5},-2) \) and \( (3-2\sqrt{5},-2) \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The foci are } (h\pm c,k)\)
\(\,\,\,\,\,\,\,\text{To solve for c use } c^2=a^2+b^2\)
\(\,\,\,\,\,\,\, c^2=16+4\)
\(\,\,\,\,\,\,\, c^2=20\)
\(\,\,\,\,\,\,\, c=\sqrt{20}\)
\(\,\,\,\,\,\,\, c=2\sqrt{5}\)
\(\,\,\,\,\,\,\,\text{The foci are } (3\pm 2\sqrt{5},-2)\)
\(\textbf{6)}\) Find the slopes of the asymptotes of the hyperbola \(\displaystyle\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1\). The slopes of the asymptotes are \( \displaystyle \pm \frac{1}{2} \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{b}{a}\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{2}{4}\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{1}{2}\)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{b}{a}\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{2}{4}\)
\(\,\,\,\,\,\,\,\text{The slopes of the asymptotes are } \pm \displaystyle\frac{1}{2}\)
\(\textbf{7)}\) Find the eccentricity of the hyperbola \(\displaystyle\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1\). The eccentricity is \( \displaystyle \frac{\sqrt{5}}{2} \)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle\frac{c}{a}\)
\(\,\,\,\,\,\,\,\text{To solve for c use } c^2=a^2+b^2\)
\(\,\,\,\,\,\,\, c^2=16+4\)
\(\,\,\,\,\,\,\, c^2=20\)
\(\,\,\,\,\,\,\, c=\sqrt{20}\)
\(\,\,\,\,\,\,\, c=2\sqrt{5}\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle \frac{2\sqrt{5}}{4}\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle \frac{\sqrt{5}}{2}\)
\(\,\,\,\,\,\,\,\text{The equation of a horizontal hyperbola is } \displaystyle\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle\frac{c}{a}\)
\(\,\,\,\,\,\,\,\text{To solve for c use } c^2=a^2+b^2\)
\(\,\,\,\,\,\,\, c^2=16+4\)
\(\,\,\,\,\,\,\, c^2=20\)
\(\,\,\,\,\,\,\, c=\sqrt{20}\)
\(\,\,\,\,\,\,\, c=2\sqrt{5}\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle \frac{2\sqrt{5}}{4}\)
\(\,\,\,\,\,\,\,\text{The eccentricity is } \displaystyle \frac{\sqrt{5}}{2}\)
\(\textbf{8)}\) Find the equation of the hyperbola with vertices \((3,4)\) and \((9,4)\) and Foci \((1,4)\) and \((11,4)\). The equation is \( \displaystyle \frac{(x-6)^2}{9}-\frac{(y-4)^2}{16}=1 \)
\(\textbf{9)}\) Use completing the square to rewrite the equation in standard form. \( x^2+4x=25y^2+250y+646 \) The equation is \( \displaystyle \frac{(x+2)^2}{25}-\frac{(y+5)^2}{1}=1 \)
\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2+4x=25y^2+250y+646\)
\(\,\,\,\,\,\,x^2+4x-25y^2-250y=646\)
\(\,\,\,\,\,\,\left(x^2+4x+\text{___}\right)-25\left(y^2+10y+\text{___}\right)=646 + \text{___} -25\left( \text{___} \right)\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\left(x^2+4x+\underline{4}\right)-25\left(y^2+10y+\underline{25}\right)=646 + \underline{4} -25\left( \underline{25} \right)\)
\(\,\,\,\,\,\,\left(x^2+4x+4\right)-25\left(y^2+10y+25\right)=646 + 4 -625\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x+2)^2-25\left(y+5\right)^2=25\)
\(\text{Step 4: Divide both sides by 25 to set equal to 1}\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+2)^2}{25}-\frac{25\left(y+5\right)^2}{25}=\frac{25}{25}\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+2)^2}{25}-\frac{\left(y+5\right)^2}{1}=1\)
\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2+4x=25y^2+250y+646\)
\(\,\,\,\,\,\,x^2+4x-25y^2-250y=646\)
\(\,\,\,\,\,\,\left(x^2+4x+\text{___}\right)-25\left(y^2+10y+\text{___}\right)=646 + \text{___} -25\left( \text{___} \right)\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\left(x^2+4x+\underline{4}\right)-25\left(y^2+10y+\underline{25}\right)=646 + \underline{4} -25\left( \underline{25} \right)\)
\(\,\,\,\,\,\,\left(x^2+4x+4\right)-25\left(y^2+10y+25\right)=646 + 4 -625\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x+2)^2-25\left(y+5\right)^2=25\)
\(\text{Step 4: Divide both sides by 25 to set equal to 1}\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+2)^2}{25}-\frac{25\left(y+5\right)^2}{25}=\frac{25}{25}\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+2)^2}{25}-\frac{\left(y+5\right)^2}{1}=1\)
See Related Pages\(\)
\(\bullet\text{ Hyperbola Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(emathhelp.net)}\)
\(\bullet\text{ Hyperbola Grapher }\)
\(\,\,\,\,\,\,\,\,\text{(Desmos.com)}\)
\(\bullet\text{ All Conic Section Notes}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Equation of a Circle}\)
\(\,\,\,\,\,\,\,\,(x-h)^2+(y-k)^2=r^2…\)
\(\bullet\text{ Parabolas}\)
\(\,\,\,\,\,\,\,\,y=a(x-h)^2+k…\)
\(\bullet\text{ Axis of Symmetry}\)
\(\,\,\,\,\,\,\,\,x=-\frac{b}{2a}…\)
\(\bullet\text{ Ellipses}\)
\(\,\,\,\,\,\,\,\,\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1…\)
\(\bullet\text{ Area of Ellipses}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\pi a b…\)
\(\bullet\text{ Hyperbolas}\)
\(\,\,\,\,\,\,\,\,\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1…\)
\(\bullet\text{ Conic Sections- Completing the Square}\)
\(\,\,\,\,\,\,\,\,x^2+8x+y^2−6y=11 \Rightarrow (x+4)^2+(y−3)^2=36…\)
\(\bullet\text{ Conic Sections- Parametric Equations}\)
\(\,\,\,\,\,\,\,\,x=h+r \cos{t}\)
\(\,\,\,\,\,\,\,\,y=k+r \sin{t}…\)
\(\bullet\text{ Degenerate Conics}\)
\(\,\,\,\,\,\,\,\,x^2−y^2=0…\)
\(\bullet\text{ Andymath Homepage}\)
In Summary
Hyperbolas are used in many different fields, including engineering, physics, and mathematics, to model and describe real-world phenomena. They are a type of geometric shape that consists of two branches, each of which is a curve that is shaped like a U or an inverted U. The branches of a hyperbola are mirror images of each other. When graphing a hyperbola it is important to identify these key parts, the vertices, the foci and the asymptotes.
Hyperbolas are conic sections. Conic secteions are curves created by the intersection of a plane and a cone. There are four types of conic sections: circles, ellipses, parabolas, and hyperbolas.
\(\cdot\)If the plane is perpendicular to the axis of the cone, the intersection of the plane and the cone will be a circle.
\(\cdot\)If the plane is between parallel and perpendicular to the axis of the cone, the intersection of the plane and the cone will be an ellipse.
\(\cdot\)If the plane is parallel to the axis of the cone, the intersection of the plane and the cone will be a parabola.
\(\cdot\)If the plane is beyond parallel to the axis of the cone, the intersection of the plane and the cone will be a hyperbola.
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