Classifying Conics

Authored by Sandra Bush

Mathematics

10th - 12th Grade

CCSS covered

Used 36+ times

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32 questions

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\left(x-3\right)^2\ -\ \frac{y^2}{25}\ =\ 1$

circle

ellipse

hyperbola

parabola

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$y\ =\ -\left(x+5\right)^2\ +\ 3$

circle

ellipse

hyperbola

parabola

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{\left(x+1\right)^2}{10}-\ \frac{\left(y-1\right)}{15}\ =\ 1$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\left(x+3\right)^2\ +\ \left(y-4\right)^2\ =\ 12$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$x^2\ +\ \left(y-6\right)^2\ =\ 1$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{\left(x-2\right)^2}{4}+\frac{\left(y-2\right)^2}{25}=\ 1$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\left(x-6\right)^2\ +\frac{y^2}{16}=1$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$x\ =\ \left(y-2\right)^2\ -\ 5$

circle

ellipse

hyperbola

parabola

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\frac{x^2}{25}-\ \frac{\left(y+2\right)^2}{9}\ =\ 1$

circle

ellipse

hyperbola

parabola

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\left(x+2\right)^2\ +\ \left(y-7\right)^2=\ 4$

circle

ellipse

hyperbola

parabola

Tags

CCSS.HSG.GPE.A.1

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Classifying Conics

(x−3)2 − y225 = 1\left(x-3\right)^2\ -\ \frac{y^2}{25}\ =\ 1(x−3)2 − 25y2​ = 1

y = −(x+5)2 + 3y\ =\ -\left(x+5\right)^2\ +\ 3y = −(x+5)2 + 3

(x+1)210− (y−1)15 = 1\frac{\left(x+1\right)^2}{10}-\ \frac{\left(y-1\right)}{15}\ =\ 110(x+1)2​− 15(y−1)​ = 1

(x+3)2 + (y−4)2 = 12\left(x+3\right)^2\ +\ \left(y-4\right)^2\ =\ 12(x+3)2 + (y−4)2 = 12

x2 + (y−6)2 = 1x^2\ +\ \left(y-6\right)^2\ =\ 1x2 + (y−6)2 = 1

(x−2)24+(y−2)225= 1\frac{\left(x-2\right)^2}{4}+\frac{\left(y-2\right)^2}{25}=\ 14(x−2)2​+25(y−2)2​= 1

(x−6)2 +y216=1\left(x-6\right)^2\ +\frac{y^2}{16}=1(x−6)2 +16y2​=1

x = (y−2)2 − 5x\ =\ \left(y-2\right)^2\ -\ 5x = (y−2)2 − 5

x225− (y+2)29 = 1\frac{x^2}{25}-\ \frac{\left(y+2\right)^2}{9}\ =\ 125x2​− 9(y+2)2​ = 1

(x+2)2 + (y−7)2= 4\left(x+2\right)^2\ +\ \left(y-7\right)^2=\ 4(x+2)2 + (y−7)2= 4

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$\left(x-3\right)^2\ -\ \frac{y^2}{25}\ =\ 1$

$y\ =\ -\left(x+5\right)^2\ +\ 3$

$\frac{\left(x+1\right)^2}{10}-\ \frac{\left(y-1\right)}{15}\ =\ 1$

$\left(x+3\right)^2\ +\ \left(y-4\right)^2\ =\ 12$

$x^2\ +\ \left(y-6\right)^2\ =\ 1$

$\frac{\left(x-2\right)^2}{4}+\frac{\left(y-2\right)^2}{25}=\ 1$

$\left(x-6\right)^2\ +\frac{y^2}{16}=1$

$x\ =\ \left(y-2\right)^2\ -\ 5$

$\frac{x^2}{25}-\ \frac{\left(y+2\right)^2}{9}\ =\ 1$

$\left(x+2\right)^2\ +\ \left(y-7\right)^2=\ 4$