Question

Find the eccentricity of the conic 25x² - 4y² = 100.
1. 5
2. \(\frac{5}{2}\)
3. \(\frac{\sqrt{29}}{2}\)
4. \(\frac{\sqrt{21}}{2}\)

Answer 1

Correct Answer - Option 3 : \(\frac{\sqrt{29}}{2}\)

Concept:

The general equation of the hyperbola is:

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

Here, coordinates of foci are (±ae, 0).

And eccentricity = \(e=\sqrt{1+\frac{b^2}{a^2}}\)

Calculation:

The equation 25x2 - 4y2 = 100 can be written as

\(\frac{x^{2}}{4}-\frac{y^{2}}{25}=1\)

This is the equation of a hyperbola.

On comparing it with the general equation of hyperbola, we get

⇒ a² = 4 and b² = 25

Now, the eccentricity is given by

\(e=\sqrt{1+\frac{25}{4}} = \frac{\sqrt{29}}{2}\)

Hence, the eccentricity is \(\frac{\sqrt{29}}{2}\).