Question

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

• A. $(\frac{5}{4},\frac{3}{4})$
• B. $(\frac{3}{4},-\frac{1}{2})$
• C. $(\frac{5}{4},\frac{3}{2})$
• D. $(\frac{3}{4},-\frac{5}{4})$

Answer 1

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)
$h=1\pm \left(\frac{\sqrt{5}}{4}\right)(-\frac{1}{\sqrt{5}})=1\mp \frac{1}{4}=\frac{3}{4},\frac{5}{4}$
$k=-1\pm \left(\frac{\sqrt{5}}{4}\right)\left(\frac{2}{\sqrt{5}}\right)=-1\pm \frac{1}{2}=-\frac{1}{2},-\frac{3}{2}$
$(\frac{3}{4},-\frac{1}{2})$ and $(\frac{5}{4},-\frac{3}{2})$ are required points.