The coordinates of the extremities of the lat | vedclass

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(B) The given equation of the parabola is $5y^2 = 4x$,which can be rewritten as $y^2 = \frac{4}{5}x$.Comparing this with the standard form $y^2 = 4ax$

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$(1/5, 2/5), (1/5, -2/5)$

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(B) The given equation of the parabola is $5y^2 = 4x$,which can be rewritten as $y^2 = \frac{4}{5}x$.Comparing this with the standard form $y^2 = 4ax$,we get $4a = \frac{4}{5}$,so $a = \frac{1}{5}$.The focus of the parabola is $(a, 0) = (\frac{1}{5}, 0)$.The latus rectum is a line perpendicular to the axis of the parabola passing through the focus,so its equation is $x = a = \frac{1}{5}$.Substituting $x = \frac{1}{5}$ into the parabola equation $y^2 = \frac{4}{5}x$,we get $y^2 = \frac{4}{5} 	imes \frac{1}{5} = \frac{4}{25}$.Thus,$y = \pm \frac{2}{5}$.Therefore,the coordinates of the extremities of the latus rectum are $(\frac{1}{5}, \frac{2}{5})$ and $(\frac{1}{5}, -\frac{2}{5})$.

The coordinates of the extremities of the latus rectum of the parabola $5y^2 = 4x$ are

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