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Geometry

Conic Sections

Concept Quizzes

Conics - Circle Standard Equation
Conics - Circle General Equation
Conics - Parabola - General
Conics - Parabola - Focus and Directrix
Conics - Ellipse - General
Conics - Ellipse - Foci
Conics - Hyperbola
Conics - Discriminant
Conic Sections - Problem Solving

Challenge Quizzes

Conic Sections: Level 2 Challenges
Conic Sections: Level 3 Challenges
Conic Sections: Level 5 Challenges

Conics - Ellipse - Foci

Suppose $F$ and $F'$ are the foci of the ellipse $\displaystyle{\frac{x^2}{121}+\frac{y^2}{64}=1}.$ If $P$ is a point on the ellipse and $\lvert \overline{FP} \rvert$ denotes the length of $\overline{FP},$ what the minimum value of $\lvert \overline{FP} \rvert^2 + \lvert \overline{F'P} \rvert^2?$

240 244 246 242

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by Brilliant Staff

The above diagram is an ellipse-shaped track. The distance between the two foci $F$ and $F'$ is $8,$ and the sum of the distances between any point $P$ on the inner side of the track and $F$ and $F'$ is $32.$ What is the maximum area of $\triangle PF'F?$

{3}\sqrt{238}

{4}\sqrt{238}

{4}\sqrt{240}

{3}\sqrt{240}

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by Brilliant Staff

For an ellipse $E$ with foci $A=(2,0)$ and $B=(10,0),$ every point $P$ on $E$ satisfies $\lvert PA \rvert + \lvert PB \rvert = 30.$ What is the equation of ellipse $E?$

Note: $\lvert PA \rvert$ denotes the length of line segment $PA$ .

\frac{(x-{6})^2}{217}+\frac{y^2}{218}=1

\frac{(x-{8})^2}{235}+\frac{y^2}{205}=1

\frac{(x-{6})^2}{225}+\frac{y^2}{209}=1

\frac{(x-{8})^2}{225}+\frac{y^2}{211}=1

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by Brilliant Staff

In the above diagram, $F$ and $F'$ are the foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $A$ and $A'$ are the end points of the major axis of the ellipse. If the area of triangle $\triangle A'PA$ is double the area of triangle $\triangle F'PF$ and the perimeter of triangle $\triangle F'PF$ is $66,$ what is the value of $a^2+b^2?$

847 845 851 849

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by Brilliant Staff

What is the equation of the circle centered at $(3,0)$ which passes through the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1?$

x^{2} + y^{2} - 6x - 7 = 0

x^{2} + y^{2} - 6x + 5 = 0

x^{2} + y^{2} - 6x - 25 = 0

x^{2} + y^{2} - 6x - 5 = 0

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by Brilliant Staff