The eccentricity of the conic frac{5}{r}=2+3 | vedclass

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(D) Given,$\frac{5}{r} = 2 + 3 \cos 	heta + 4 \sin 	heta$. Divide by $2$: $\frac{5/2}{r} = 1 + \frac{3}{2} \cos 	heta + 2 \sin 	heta$. We can writ

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$\frac{5}{2}$

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(D) Given,$\frac{5}{r} = 2 + 3 \cos 	heta + 4 \sin 	heta$. Divide by $2$: $\frac{5/2}{r} = 1 + \frac{3}{2} \cos 	heta + 2 \sin 	heta$. We can write $\frac{3}{2} \cos 	heta + 2 \sin 	heta$ in the form $e \cos(	heta - \phi)$,where $e = \sqrt{(\frac{3}{2})^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$. Thus,the equation becomes $\frac{5/2}{r} = 1 + \frac{5}{2} \cos(	heta - \phi)$. Comparing this with the standard polar form of a conic $\frac{l}{r} = 1 + e \cos(	heta - \phi)$,we get the eccentricity $e = \frac{5}{2}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

The eccentricity of the conic $\frac{5}{r}=2+3 \cos \theta+4 \sin \theta$ is

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