Q119 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$. Then its equation is
- Correct answerA$x^2-y^2=32$
- B$\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$
- C$3 x^2-2 y^2=7$
- D$2 x^2-3 y^2=7$
Solution
The distance between the foci of a hyperbola is given by $2ae = 16$.
Given the eccentricity $e = \sqrt{2}$, we have $2a\sqrt{2} = 16$, which simplifies to $a\sqrt{2} = 8$, so $a = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$.
Squaring $a$, we get $a^2 = (4\sqrt{2})^2 = 16 \times 2 = 32$.
For a hyperbola, the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.
Substituting the known values, $b^2 = 32((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
The standard equation of a hyperbola centered at the origin is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Substituting $a^2 = 32$ and $b^2 = 32$, we get $\dfrac{x^2}{32} - \dfrac{y^2}{32} = 1$, which simplifies to $x^2 - y^2 = 32$.
Answer: $x^2-y^2=32$
Given the eccentricity $e = \sqrt{2}$, we have $2a\sqrt{2} = 16$, which simplifies to $a\sqrt{2} = 8$, so $a = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$.
Squaring $a$, we get $a^2 = (4\sqrt{2})^2 = 16 \times 2 = 32$.
For a hyperbola, the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1)$.
Substituting the known values, $b^2 = 32((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
The standard equation of a hyperbola centered at the origin is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Substituting $a^2 = 32$ and $b^2 = 32$, we get $\dfrac{x^2}{32} - \dfrac{y^2}{32} = 1$, which simplifies to $x^2 - y^2 = 32$.
Answer: $x^2-y^2=32$
Hence, the correct answer is option: 'A'.