Conics - Hyperbola Practice Problems Online

Consider a branch of the hyperbola $x^2-2y^2-2\sqrt{5}x-4\sqrt{2}y-3=0$ with vertex at a point $A.$ Also, let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to point $A,$ what is the area of triangle $ABC ?$

What are the coordinates of the foci of the rectangular hyperbola $x^2-y^2=7 ^2?$

(0,{7}\sqrt{2}) \text{ and } (0,-{7}\sqrt{2})

({7}\sqrt{3},0) \text{ and } (-{7}\sqrt{3},0)

(0,{7}\sqrt{3}) \text{ and } (0,-{7}\sqrt{3})

({7}\sqrt{2},0) \text{ and } (-{7}\sqrt{2},0)

What are the coordinates of the vertices of the hyperbola $4x^2-9y^2-40x+54y -17=0 ?$

(3,6) \text{ and } (-3,6)

(3,0) \text{ and } (-3,0)

(8,6) \text{ and } (2,6)

(8,3) \text{ and } (2,3)

Given points $F=(5,0)$ and $F'=(-5,0),$ what is the equation of the locus of points $P$ in a plane such that $\left\lvert \left\lvert \overline{PF} \right\rvert - \left\lvert \overline{PF'} \right\rvert \right\rvert= 8 ?$

Note: $\lvert \overline{AB} \rvert$ denotes the length of line segment $\overline{AB}.$

\frac{x^2}{16}-\frac{y^2}{9}=1

\frac{x^2}{16}-\frac{y^2}{9}=-1

\frac{x^2}{9}-\frac{y^2}{16}=1

\frac{x^2}{9}-\frac{y^2}{16}=-1

Suppose that $(7,12)$ and $(11,8)$ are the foci of a hyperbola passing through the origin. Then the eccentricity of the hyperbola is $\text{\_\_\_\_\_\_\_\_\_}.$

\frac{\sqrt{32}}{\sqrt{193}+\sqrt{185}}

\frac{\sqrt{32}}{\sqrt{193}-\sqrt{185}}

\frac{\sqrt{64}}{\sqrt{193}-\sqrt{185}}

\frac{\sqrt{32}}{2(\sqrt{193}+\sqrt{185})}

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Conics - Hyperbola

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