Question

Find the directrix of the given parabola $4y^2-6x-4y=5$

• A. $x=-\frac{11}{8}$
• B. $x=-\frac{11}{7}$
• C. $x=-\frac{22}{7}$
• D. $x=-\frac{9}{8}$

Answer 1

The correct option is A $x=-\frac{11}{8}$

$4y^2-6x-4y=5$
or $4(y^2-y)=6x+5.$

Make perfect square.

$4(y^2-y+\frac{1}{4})=6x+5+4.\frac{1}{4}=6(x+1)$

or ${(y-\frac{1}{2})}^2=\frac{3}{2}.(x+1).$
Above is of the form $Y^2=4AX.$

where $Y=y-1/2,X=x+1,4A=3/2$

$\therefore A=3/8.$

Vertex : $X=0,Y=0$ or $x+1=0,y-1/2=0$
or $(-1,1/2)$

Focus : $X=A,Y=0$ or $x+1=3/8,y-1/2=0$
or $(-5/8,1/2)$

Directrix : $X=-A$ or $x+1=-3/8$
or $x=-11/8$

L.R. : $X=A$ or $x+1=3/8$ or $x=-5/8$

and length $4A=3/2$

Axis : $Y=0$ or $y-1/2=0$
or $y=1/2$

T.V. : $X=0$ or $x+1=0$
or $x=-1$
Ans: A