Question types

Hyperbola question types

64 questions across 3 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

64
Questions
3
Question groups
5
Question types
Sample Questions

Hyperbola questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Equation of the hyperbola whose vertices are $(\pm3,0)$ and foci at $(\pm5,0),$ is
  1. 16x2 - 9y2 = 144
  2. 9x2 - 16y2 = 144
  3. 25x2 - 9y2 = 225
  4. 9x2 - 25y2 = 81
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If e1 and e2 are respectively the eccentricities of the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1$ and the hyperbola $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ then the relation between e1 and e2 is
  1. $3\text{e}_1^2 + \text{e}_2^2 = 2$
  2. $\text{e}_1^2 + 2\text{e}_2^2 = 3$
  3. $2\text{e}_1^2 +\text{e}_2^2 = 3$
  4. $\text{e}_1^2 + 3\text{e}_2^2 = 2$
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The equation of the conic with focus at (1, -1) directrix along x - y + 1 = 0 and eccentricity $\sqrt2$ is
  1. xy = 1
  2. 2xy + 4x - 4y - 1 = 0
  3. x2 - y2 = 1
  4. 2xy - 4x + 4y + 1 = 0
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If the foci of the ellipse $\frac{\text{x}^{2}}{16}+\frac{\text{y}^{2}}{\text{b}^{2}}=1$ and the hyperbola $\frac{\text{x}^{2}}{144}-\frac{\text{y}^{2}}{81}=\frac{1}{25}$ coincide, value of $\text{b}^{2}$
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If e and e are respectively the eccentricities of the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1$ and the hyperbola $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ then write the value of $2\text{e}_1^2 +\text{e}_2^2.$
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Find the eccentricity, coordinates of the foci, equations of directrices and lenght of the latus-rectum of the hyperbola 
$2\text{x}^{2}-3\text{y}^{2}=5.$
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In each the following find the equation of the hyperbola satisfying the given conditions:

Foci $(\pm3\sqrt{5}, 0)$, the latus-rectum = 8 [NCERT]

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