The foci of a hyperbola coincide with the foci of the ellipse \(\rm \frac{x^2}{25}+\frac{y^2}{16}=1\), find the equation of the hyperbola if its eccentricity is 3.
1. \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\)
2. \(\rm \frac{x^2}{8}-\frac{y^2}{1}=1\)
3. \(\rm \frac{x^2}{9}-\frac{y^2}{16}=1\)
4. \(\rm \frac{x^2}{25}-\frac{y^2}{16}=1\)
Correct Answer – Option 1 : \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\)
Concept:
The distance between a focus of the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and its center is denoted by c and is equal to c2 = a2 – b2.
The distance between a focus of a hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its center is denoted by c and is equal to c2 = a2 + b2, and its eccentricity e is equal to \(\rm \frac{c}{a}\).
Calculation:
The distance between the focus and the center of the ellipse \(\rm \frac{x^2}{25}+\frac{y^2}{16}=1\) is c2 = 25 – 16 = 9
⇒ c = 3.
Since the hyperbola has the same foci, its c will be the same as that of the ellipse.
Using e = \(\rm \frac{c}{a}\) for the hyperbola, we get:
3 = \(\rm \frac{3}{a}\)
⇒ a = 1
Using c2 = a2 + b2, we get:
32 = 12 + b2
⇒ b2 = 8.
The equation of the hyperbola is \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\).