Question

Which of the following in INCORRECT about the parabola $x^2+4y+3x=2$

• A. Its vertex is at $(-\frac{3}{2},\frac{17}{16})$
• B. Its focus is at $(-\frac{3}{2},\frac{1}{16})$
• C. The directrix is $y=-\frac{1}{16}$
• D. The axis is $x=-\frac{3}{2}$

Answer 1

Answer: (C)

The directrix is $y=-\frac{1}{16}$

Rewrite the given equation as following :
$x^2+3x=-4y+2\Rightarrow x^2+3x+\frac{9}{4}=-4y+2+\frac{9}{4}$
$\Rightarrow {(x+\frac{3}{2})}^2=-4(y-\frac{17}{16})$ .......... (i)
Shifting the origin to $A(-\frac{3}{2},\frac{17}{16})$, we substitute $x+\frac{3}{2}=X$ and $y-\frac{17}{16}=Y$ ........... (ii)
The equation (i) becomes, $X^2=-4Y$ ........ (iii)
It is a downward parabola with $a=-1$, Vertex $(0,0)$, focus $(0,-1)$ directrix $Y=1$ and axis $X=0$.
Substituting from (ii), we get in the original coordinate system. Vertex $(0,0)$ i.e., $X$ $=$ $0$ & $Y$ $=$ $0$.
$\therefore x+\frac{3}{2}$ and $y-\frac{17}{16}=0\Rightarrow x=-\frac{3}{2},y=\frac{17}{16}$
The vertex is $(-\frac{3}{2},\frac{17}{16})$. Focus $(0,-1)$ i.e. $X$ $=$ $0$ & $Y$ $=$ $-1$.
$\therefore x+\frac{3}{2}=0$ and $y-\frac{17}{16}=-1\Rightarrow x=-\frac{3}{2},y=\frac{1}{16}$
The focus is $(-\frac{3}{2},\frac{1}{16})$
Directrix $Y=1,\Rightarrow y-\frac{17}{16}=1\Rightarrow y=\frac{33}{16}$
Axis $X=0\Rightarrow x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}$