Question

Find the equation of the parabola whose axis is parallel to $y$ - axis and passes through the points $(0,4),(1,9)$ and $(-2,6).$

• A. ${(x+\frac{3}{4})}^2=4\frac{1}{8}(y-\frac{23}{8})$
• B. ${(x-\frac{3}{4})}^2=2\frac{1}{8}(y-\frac{23}{8})$
• C. ${(x+\frac{4}{3})}^2=3\frac{1}{8}(y+\frac{23}{8})$
• D. ${(x+\frac{4}{3})}^2=4\frac{1}{8}(y-\frac{23}{8})$

Answer 1

The correct option is A ${(x+\frac{3}{4})}^2=4\frac{1}{8}(y-\frac{23}{8})$

Let the vertex of the parabola be the point $(h,k)$ and length of its latus rectum be $4a$.

Since its axis is parallel to $y$ - axis, its equation can be written as
$(x-h{)}^2=4a(y-k)$ ..... $\left(1\right)$
It passes through the given points $(0,4),(1,9)$ and $(-2,6)$
$\therefore {(0-h)}^2=4a(4-k)\Rightarrow h^2=4a(4-k)$ ...... $\left(2\right)$

${(1-h)}^2=4a(9-k)$ $\Rightarrow 1-2h+h^2=4a(9-k)$ ...... $\left(3\right)$

${(-2-h)}^2=4a(6-k)$ $\Rightarrow 4+4h+h^2=4a(6-k)$ ...... $\left(4\right)$

Subtracting $\left(2\right),\left(3\right)$ and $\left(3\right),\left(4\right)$ we have

$1-2h=20a$ ..... $\left(5\right)$ and $3+6h=-12a$ i.e. $1+2h=-4a$ ..... $\left(6\right)$

Then solving $\left(5\right)$ and $\left(6\right)$, we get
$a=\frac{1}{8}$ and $h=-\frac{3}{4}$

Substituting in any of the equations $\left(2\right),\left(3\right)$ and $\left(4\right)$, we get

$k=\frac{23}{8}.$

Substitutingin $\left(1\right),$ the equation of parabola is
${(x+\frac{3}{4})}^2=4\frac{1}{8}(y-\frac{23}{8})$
Its vertex is the point $(-\frac{3}{4},\frac{23}{8})$ and L.R.$=\frac{1}{2}.$
Ans: A