Question

The equation of the parabola, whose vertex is $\left( -1,-2 \right)$, axis is vertical and which passes through the point $\left( 3,6 \right)$ is
A. ${{x}^{2}}+2x-2y-3=0$
B. $2{{x}^{2}}=3y$
C. ${{x}^{2}}-2x+y+3=0$
D. None of these

Answer 1

To find the required equation of parabola whose vertex is given, we will use the given condition that the axis is vertical. So, the equation of the parabola would be ${{\left( x-A \right)}^{2}}=4a\left( y-B \right)$, where $A$ and $B$ are vertices of the parabola. Now, we will substitute the value of the vertex and use the second condition to find the value of $a$ that the parabola passes through the point $\left( 3,6 \right)$. So, this point satisfies the obtained equation. So, we will substitute the value of $x$ and $y$ with this point and evaluate the value of $a$. After substituting the value of $a$ in the obtained equation, we will get the equation of required parabola.

Complete step by step solution:
According to the question, the axis of parabola is vertical. Then the required equation should be as:
$\Rightarrow {{\left( x-A \right)}^{2}}=4a\left( y-B \right)$
Where $A$ and $B$ are coordinates of the vertex of the parabola.
Since, we already have the points of the vertex as $\left( -1,-2 \right)$. So, we will substitute their values accordingly as $-1$ for $A$ and $-2$ for $B$. The equation will be:
$\Rightarrow {{\left( x-\left( -1 \right) \right)}^{2}}=4a\left( y-\left( -2 \right) \right)$
Simplify the above equation as:
$\Rightarrow {{\left( x+1 \right)}^{2}}=4a\left( y+2 \right)$ … $\left( i \right)$
Now, we will use the second condition to calculate the value of $a$ so that we can get the required equation. Since, the given point $\left( 3,6 \right)$ lies on the parabola. So, this point satisfies the equation of parabola as:
$\Rightarrow {{\left( 3+1 \right)}^{2}}=4a\left( 6+2 \right)$
Complete the operation within bracket as:
$\Rightarrow {{\left( 4 \right)}^{2}}=4a\left( 8 \right)$
Now, we will do the square left side and multiplication right side as:
$\Rightarrow 16=32a$
It can be written as:
$\Rightarrow a=\dfrac{16}{32}$
After simplifying it into simplest form, we will have:
$\Rightarrow a=\dfrac{1}{2}$
Now, we will substitute the value of $a$in equation $\left( i \right)$ as:
$\Rightarrow {{\left( x+1 \right)}^{2}}=4\times \dfrac{1}{2}\left( y+2 \right)$
Here, we will simplify the equation as:
$\begin{aligned} & \Rightarrow {{x}^{2}}+2x+1=2\left( y+2 \right) \\\ & \Rightarrow {{x}^{2}}+2x+1=2y+4 \\\ \end{aligned}$
Now, we will move all the variables and number one side as:
$\Rightarrow {{x}^{2}}+2x+1-2y-4=0$
After solving the above equation, we will have:
$\Rightarrow {{x}^{2}}+2x-2y-3=0$
Hence, option $a$ is the correction answer.

Note: We can check the solutions if it is correct or not by using the point $\left( 3,6 \right)$ because this point will satisfy the equation.
Since, we obtained the required equation of parabola that is:
$\Rightarrow {{x}^{2}}+2x-2y-3=0$
Take L.H.S.
$\Rightarrow {{x}^{2}}+2x-2y-3$
Now, substitute the value of $x$ and $y$ with the point $\left( 3,6 \right)$ as:
$\Rightarrow {{3}^{2}}+2\times 3-2\times 6-3$
Here, we will simplify the equation as:

& \Rightarrow 9+6-12-3 \\\ & \Rightarrow 15-15 \\\ & \Rightarrow 0 \\\ \end{aligned}$$ That is equal to R.H.S. Hence, the solution is correct.