Consider a curve $(2 x + y - 1)^{2} = 5 (x - 2 y - 3)$ then the possible coordinates of foci are

(\frac{5}{4}, \frac{3}{4})

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(\frac{3}{4}, - \frac{1}{2})

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(\frac{5}{4}, + \frac{3}{2})

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(\frac{3}{4}, \frac{5}{4})

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Solution

The correct option is B $(\frac{3}{4}, - \frac{1}{2})$
2x + y – 1= 0 is the axis and (1, - 1) is the point of intersection with the tangent at the vetex, length of latus rectum is $\sqrt{5}$
So, coordinates can be (h,k)
$h = 1 \pm (\frac{\sqrt{5}}{4}) (- \frac{1}{\sqrt{5}}) = 1 \mp \frac{1}{4} = \frac{3}{4}, \frac{5}{4} k = - 1 \pm (\frac{\sqrt{5}}{4}) (\frac{2}{\sqrt{5}}) = - 1 \pm \frac{1}{2} = - \frac{1}{2}, - \frac{3}{2} (\frac{3}{2}, - \frac{1}{2}) a n d (\frac{5}{4}, - \frac{3}{2})$
are required points

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Consider a curve (2x+y−1)2=5(x−2y−3) then the possible coordinates of foci are

Consider a curve $(2 x + y - 1)^{2} = 5 (x - 2 y - 3)$ then the possible coordinates of foci are