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.

Q 1MCQ1 Mark
Equation of the hyperbola whose vertices are $( \pm 3,0)$ and foci at $( \pm 5,0)$, is
  • $16 x^2-9 y^2=144$
  • B
    $9 x^2-16 y^2=144$
  • C
    $25 \mathrm{x}^2-9 \mathrm{y}^2=225$
  • D
    $9 x^2-25 y^2=81$

Answer: A.

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Q 2MCQ1 Mark
If $e_1$ and $e_2$ 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 $e_1$ and $e_2$ is
  • A
    $3\text{e}_1^2 + \text{e}_2^2 = 2$
  • B
    $\text{e}_1^2 + 2\text{e}_2^2 = 3$
  • $2\text{e}_1^2 +\text{e}_2^2 = 3$
  • D
    $\text{e}_1^2 + 3\text{e}_2^2 = 2$

Answer: C.

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Q 3MCQ1 Mark
The eccentricity of the hyperbola $x^2 - 4y^2 = 1$
  • A
    $\frac{\sqrt3}{2}$
  • ${\frac{\sqrt5}{2}}$
  • C
    ${\frac{2}{\sqrt3}}$
  • D
    $\frac{2}{\sqrt5}$

Answer: B.

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Q 4MCQ1 Mark
The distance between the directrices of the hyperbola $\text{x}=8\sec\theta,\text{y}=8,$ is
  • $8\sqrt2$
  • B
    $16\sqrt2$
  • C
    $4\sqrt2$
  • D
    $6\sqrt2$

Answer: A.

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Q 5MCQ1 Mark
The equation of the conic with focus at $(1, -1)$ directrix along $x - y + 1 = 0$ and eccentricity $\sqrt2$ is
  • A
    $x y=1$
  • B
    $2 x y+4 x-4 y-1=0$
  • C
    $x^2-y^2=1$
  • $2 x y-4 x+4 y+1=0$

Answer: D.

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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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