Question

If focus of parabola is at $(0,-3)$ and its directrix is $y=3$ then its equation is:
A.${{x}^{2}}=12y$
B.${{x}^{2}}=-12y$
C.${{y}^{2}}=-12x$
D.${{y}^{2}}=12x$

Answer 1

Hint: Here, we have to apply the distance formula, i.e. if $A=\left( {{x}_{1}},{{y}_{1}} \right)$ and $B=\left( {{x}_{2}},{{y}_{2}} \right)$ then the distance between $A$ and $B$ is $AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$. And also apply that any point on the parabola is equidistant from the focus and the directrix. i.e. If $P(a,b)$ is the point and $S(0,-3)$the focus, $M$the directrix, then $SP=PM$

Complete step-by-step answer:
Given that the focus of the parabola is $S(0,-3)$ and directrix is $y=3$.
Here, we have to find the equation of the parabola.
For that first consider $P(a,b)$ be any point on the parabola.
Next, we have to calculate the distance of the point $P(a,b)$ from the focus $S(0,-3)$.
For that we have the distance formula, i.e. if $A=\left( {{x}_{1}},{{y}_{1}} \right)$ and $B=\left( {{x}_{2}},{{y}_{2}} \right)$ then the distance between $A$ and $B$ is given by the formula:
$AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Here, $\left( {{x}_{1}},{{y}_{1}} \right)=(a,b)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=(0,-3)$. By applying distance formula we get:
$\begin{aligned} & SP=\sqrt{{{\left( a-0 \right)}^{2}}+{{\left( b-(-3) \right)}^{2}}} \\\ & SP=\sqrt{{{a}^{2}}+{{\left( b+3 \right)}^{2}}}\text{ }.....\text{ (1)} \\\ \end{aligned}$
Next, we have to find the distance of the point $P(a,b)$ from the directrix M, $y=3$.
We know that the distance of the point $({{x}_{0}},{{y}_{0}})$ from the line $y=c$ is $\left| {{y}_{0}}-c \right|$.
Here, we have $\left( {{x}_{0}},{{y}_{0}} \right)=(a,b)$, $c=3$
i.e. we can write:
$PM=\left| b-3 \right|\text{ }.......\text{ (2) }$
We know that any point on the parabola is equidistant from the focus and the directrix.
Therefore, we can say that:
$SP=PM$
By substituting equation (1) and (2) in the above equation we get:
$\sqrt{{{a}^{2}}+{{\left( b+3 \right)}^{2}}}=\left| b-3 \right|$
Next, by squaring equation on both the sides we obtain:
$\begin{aligned} & {{\left( \sqrt{{{a}^{2}}+{{\left( b+3 \right)}^{2}}} \right)}^{2}}={{\left| b-3 \right|}^{2}} \\\ & {{a}^{2}}+{{\left( b+3 \right)}^{2}}={{(b-3)}^{2}} \\\ \end{aligned}$
We also know that ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, our equation becomes:
$\begin{aligned} & {{a}^{2}}+({{b}^{2}}+2\times 3\times b+{{3}^{2}})={{b}^{2}}-2\times b\times 3+{{3}^{2}} \\\ & {{a}^{2}}+({{b}^{2}}+6b+9)={{b}^{2}}-6b+9 \\\ & {{a}^{2}}+{{b}^{2}}+6b+9={{b}^{2}}-6b+9 \\\ \end{aligned}$
Next, we have to bring all terms to the left side. Then the sign changes, therefore our equation becomes:
${{a}^{2}}+{{b}^{2}}+6b+9-{{b}^{2}}+6b-9=0.$
Now, by simplification we obtain:
${{a}^{2}}+12b=0$
Next, by taking $12b$ to the right side it becomes $-12b$. Hence, our equation yields to:
${{a}^{2}}=-12b$
Now, by substituting $a=x$ and $b=y$ we get the equation:
${{x}^{2}}=-12y$
Hence, the equation of parabola with focus $(0,-3)$ and directrix $y=3$ is ${{x}^{2}}=-12y$
Therefore, the correct answer for this question is option (b).

Note: Here, instead of taking the point (a,b)we can take as (x,y)and solve directly which will help to save your time. Also you should know that every point on the parabola is equidistant from the focus and directrix. Only with this knowledge, we can equate those distances.