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CONIC SECTION - HYPERBOLA

Authored by Nilesh gulati

Mathematics

11th Grade

CCSS covered

Used 65+ times

CONIC SECTION - HYPERBOLA
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7 questions

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1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Determine the length of the transverse axis

1.5

2

3

4

Tags

CCSS.HSG.GPE.A.3

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Determine the equation of the hyperbola.

x24y2=1\frac{x^2}{4}-y^2=1

x22y2=1\frac{x^2}{2}-y^2=1

x24+y2=1\frac{x^2}{4}+y^2=1

x22+y2=1\frac{x^2}{2}+y^2=1

Tags

CCSS.HSG.GPE.A.3

CCSS.HSA.CED.A.2

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

What is the equation describing the relationship between dimensions and foci of a hyperbola?

c2 = a2 + b2
c2 = a2 - b2
4p
(x-h)2 + (y-k)2 = r2

Tags

CCSS.HSG.GPE.A.3

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Is the hyperbola vertical or horizontal? 

Vertical
Horizontal

Tags

CCSS.HSG.GPE.A.3

5.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Coordinates of the foci of the hyperbola

 x29y216=1\frac{x^2}{9}-\frac{y^2}{16}=1  are

 (± 3, 0)\left(\pm\ 3,\ 0\right)  

 (± 4, 0)\left(\pm\ 4,\ 0\right)  

 (± 5, 0)\left(\pm\ 5,\ 0\right)  

none of these

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Equation of the hyperbola with vertices

 (0, ± 5)\left(0,\ \pm\ 5\right)  and foci  (0, ± 8)\left(0,\ \pm\ 8\right)  is

 y225x239=1\frac{y^2}{25}-\frac{x^2}{39}=1  

 x225y239=1\frac{x^2}{25}-\frac{y^2}{39}=1  

 x225+y239=1\frac{x^2}{25}+\frac{y^2}{39}=1  

none of these

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Equation of hyperbola if length of transverse axis is 8 and foci \left(\pm\ 5,\ 0\right)  

 x216y29=1\frac{x^2}{16}-\frac{y^2}{9}=1  

 x29y216=1\frac{x^2}{9}-\frac{y^2}{16}=1  

 y216x29=1\frac{y^2}{16}-\frac{x^2}{9}=1  

none of these

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