Question

Find the vertex, axis, focus, directrix, latus rectum of the parabola $4y^2+12x-20y+67=0$.

• A. Vertex : $(-\frac{7}{2},\frac{5}{2})$, Axis : $y=\frac{5}{2}$, Focus : $(-\frac{17}{4},\frac{5}{2})$, Directrix :$x=-\frac{11}{4}$; LR $=3$
• B. Vertex : $(-\frac{9}{2},\frac{5}{2})$, Axis : $y=\frac{5}{2}$, Focus : $(-\frac{17}{4},\frac{5}{2})$, Directrix :$x=-\frac{11}{4}$; LR $=4$
• C. Vertex : $(-\frac{9}{2},\frac{5}{2})$, Axis : $y=\frac{5}{2}$, Focus : $(-\frac{17}{4},\frac{5}{2})$, Directrix :$x=-\frac{11}{4}$; LR $=3$
• D. Vertex : $(-\frac{7}{2},\frac{5}{2})$, Axis : $y=\frac{5}{2}$, Focus : $(-\frac{17}{4},\frac{5}{2})$, Directrix :$x=-\frac{11}{4}$; LR $=4$

Answer 1

The correct option is A Vertex : $(-\frac{7}{2},\frac{5}{2})$, Axis : $y=\frac{5}{2}$, Focus : $(-\frac{17}{4},\frac{5}{2})$, Directrix :$x=-\frac{11}{4}$; LR $=3$

Consider, $4y^2+12x-20y+67=0$

$4(y^2-5y)=-12x-67$

$4(y^2-2.\frac{5}{2}y+\frac{5^2}{2^2})=-12x-67+25=-12x-42$

${(y-\frac{5}{2})}^2=-3(x+\frac{7}{2})$

$\Rightarrow$ Vertex $:(-\frac{7}{2},\frac{5}{2})$,

Axis $:y=\frac{5}{2}$

Focus $:(-\frac{7}{2}-\frac{3}{4},\frac{5}{2})=(-\frac{17}{4},\frac{5}{2})$

Directrix $:x+\frac{7}{2}=\frac{3}{4}\Rightarrow x=-\frac{11}{4}$

And Latus rectum $=4\times \frac{3}{4}=3$