Question

The length of the latus-rectum of the parabola $4y^2+2x-20y+17=0$

• A. 3
• B. 6
• C. $\frac{1}{2}$
• D. 9

Answer 1

The correct option is C

$\frac{1}{2}$

Given: $4y^2+2x-20y+17=0$

$\Rightarrow y^2+\frac{x}{2}-5y+\frac{17}{4}=0$

$\Rightarrow {(y-\frac{5}{2})}^2+\frac{x}{2}-1=0$

$\Rightarrow {(y-\frac{5}{2})}^2=-1(\frac{x}{2}-2)$

$\Rightarrow {(y-\frac{5}{2})}^2=\frac{-1}{2}(x-4)$

$LetX=x-4,Y=y-\frac{5}{2}$

$\therefore Y^2=\frac{-X}{2}$

$\therefore \text{Length of the latus rectum}=4a=\frac{1}{2}units.$