Question

Find the eccentricity and length of latus rectum for the hyperbola x² - y² = 1.
1. 1, √2
2. √2, 2
3. 2, 1
4. √2, 1

Answer 1

Correct Answer - Option 2 : √2, 2

Concept:

The standard equation of a hyperbola:

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

Where 2a and 2b are the length of the transverse axis and conjugate axis respectively and center (0, 0)

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2+b^2}\)

Calculation:

Given hyperbola x2 - y2 = 1

Compare with standard equation of  hyperbola,

So, a² = 1, b² = 1

The eccentricity e = \(\rm \sqrt{a^2+b^2}\over a\)

⇒ e = \(\rm \sqrt{1^2+1^2}\over 1\)

⇒ e = \( \boldsymbol {\rm\sqrt{2}}\)

Now length of latus recta (l)= \(\rm 2b^2 \over a\)

⇒ l = \(\rm 2\times{1^2 \over 1}\)

⇒ l = \( \boldsymbol {\rm2}\)