Question

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

Answer 1

Hint: The general form of the parabola $y=a{{x}^{2}}+bx+c$ is to be used while solving this question.

Complete step-by-step answer:
In the question, it is given that the axis of the parabola is parallel to the $y-$axis. So, the parabola would look like the figure below.

We know that the general formula for this form of parabola is given by,
$y=a{{x}^{2}}+bx+c\ldots \ldots \ldots \left( i \right)$
It is given in the question that the parabola passes through the points $\left( 0,2 \right),\left( -1,0 \right)$ and $\left( 1,6 \right)$. Since the parabola passes through the three points, we know that these points must satisfy the equation of the parabola. So, we can substitute each point in the equation $\left( i \right)$ and formulate three sets of equations.
Considering the first point $\left( 0,2 \right)$ and substituting the values of $x=0,y=2$ in equation $\left( i \right)$, we get

& 2=a\times 0+b\times 0+c \\\ & c=2\ldots \ldots \ldots \left( ii \right) \\\ \end{aligned}$$ Considering the second point $$\left( -1,0 \right)$$ and substituting the values of $$x=-1,y=0$$ in equation $\left( i \right)$, we get $$\begin{aligned} & 0=a\times {{\left( -1 \right)}^{2}}+b\times -1+c \\\ & 0=a-b+c \\\ \end{aligned}$$ Now substituting the value of $$c$$ from equation $\left( ii \right)$ in the above equation, we get $$\begin{aligned} & 0=a-b+2 \\\ & b-a=2\ldots \ldots \ldots \left( iii \right) \\\ \end{aligned}$$ Considering the last point $$\left( 1,6 \right)$$ and substituting the values of $$x=1,y=6$$ in equation $\left( i \right)$, we get $\begin{aligned} & 6=a\times {{1}^{2}}+b\times 1+c \\\ & 6=a+b+c \\\ \end{aligned}$ Now substituting the value of $$c$$ from equation $\left( ii \right)$ in the above equation, we get $$\begin{aligned} & 6=a+b+2 \\\ & a+b=4\ldots \ldots \ldots \left( iv \right) \\\ \end{aligned}$$ We have two equations $$\left( iii \right)$$ and $\left( iv \right)$ to get the values of the $$a$$ and $b$. So, adding the equations, $$\dfrac{\begin{aligned} & b-a=2 \\\ & a+b=4 \\\ \end{aligned}}{\begin{aligned} & 2b=6 \\\ & b=3 \\\ \end{aligned}}$$ Substituting $$b=3$$ in equation $$\left( iii \right)$$, we get $$\begin{aligned} & 3-a=2 \\\ & a=1 \\\ \end{aligned}$$ Now we have the values as $a=1,b=3,c=2$. So, we can substitute this in equation $\left( i \right)$, $$y={{x}^{2}}+3x+2$$ Therefore, the required equation of the parabola is obtained as $$y={{x}^{2}}+3x+2$$. Note: As three points are given in the question, we can formulate three equations and easily compute the three unknowns in the equation. If you are familiar with the cross-multiplication method, you can solve the equations and get the values of $$a,b,c$$ in less time.