The y-coordinates of the extremities of latus rectum of hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
are $\frac{\pm b^{2}}{a}$

True

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False

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Solution

The correct option is A True

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Similar questions

b^{2} x^{2} + y^{2} = a^{2} b^{2}

b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2} is (b \in R)

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

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

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

x = \pm \frac{a}{e} .

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