The foci of a hyperbola coincide with the foci of the ellipse \(\rm \f

1.	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. \(\rm \frac{x^2}{25}-\frac{y^2}{16}=1\)
Answer» 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 \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\).

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. \(\rm \frac{x^2}{25}-\frac{y^2}{16}=1\)

Answer» 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 \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\).

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