Question

Focus of the parabola $4x^2-12x+8y+13=0$ is

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

Answer 1

The correct option is A $(\frac{-1}{2},1)$

$4x^2-12x+8y+13=0\Rightarrow 4(x^2-3x)+13+8y=0\Rightarrow 4[x^2-\frac{3}{2}-2x+\frac{9}{4}-\frac{9}{4}]+13+8y=0\Rightarrow 4(x-\frac{3}{2}{)}^2-9+13+8y=0\Rightarrow 4(x-\frac{3}{2}{)}^2+8y+4=0\Rightarrow (x-\frac{3}{2}{)}^2+8y+4=0\Rightarrow (x-\frac{3}{2}{)}^2\times 4=-(8y+4)\Rightarrow (x-\frac{3}{2}{)}^2=-(2y+1)\Rightarrow (x-\frac{3}{2}{)}^2=-2(y+\frac{1}{2})$

$\therefore$ focus is $(a+h,k)$

Here $4a=-2\Rightarrow a=-\frac{1}{2}h=\frac{3}{2},k=\frac{-1}{2}$

$\therefore$ focus is $(\frac{-1}{2}+\frac{3}{2},\frac{-1}{2})=(1,\frac{-1}{2})$