The length of the latus rectum of the hyperbo | vedclass

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(A) The given equation of the hyperbola is $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$. Dividing by $4$,we get $\frac{x^2}{4\cos^2 \al

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$2\left| \frac{1 - \cos 2\alpha}{\cos \alpha} \right|$

Vedclass · Answer

(A) The given equation of the hyperbola is $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$. Dividing by $4$,we get $\frac{x^2}{4\cos^2 \alpha} - \frac{y^2}{4\sin^2 \alpha} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 4\cos^2 \alpha$ and $b^2 = 4\sin^2 \alpha$. Thus,$a = 2|\cos \alpha|$ and $b = 2|\sin \alpha|$. The length of the latus rectum is given by $LR = \frac{2b^2}{a}$. Substituting the values,$LR = \frac{2(4\sin^2 \alpha)}{2|\cos \alpha|} = \frac{4\sin^2 \alpha}{|\cos \alpha|}$. Using the identity $2\sin^2 \alpha = 1 - \cos 2\alpha$,we get $LR = \frac{2(1 - \cos 2\alpha)}{|\cos \alpha|} = 2\left| \frac{1 - \cos 2\alpha}{\cos \alpha} \right|$.

The length of the latus rectum of the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$ is (where $\alpha \neq \frac{n\pi}{2}, n \in I$).

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