What is the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through the point $\left( 12,-4 \right)$ ?
Answer
545.7k+ views
Hint: To find the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through point $\left( 12,-4 \right)$ , we have to consider the vertex form of the parabola which is given by $y=a{{\left( x-h \right)}^{2}}+k$ , where “a” is the distance from the origin to the focus and $\left( h,k \right)$ is the vertex. We have to substitute the given vertex in the above formula. Then, we have to find the value of a from the given point by substituting the point in the vertex form. Then, we have to solve for a and substitute this value of a in the vertex form which we obtained after substituting the vertex.
Complete step-by-step solution:
We have to find the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through point $\left( 12,-4 \right)$ . We know that the equation of a parabola with vertex $\left( h,k \right)$ is given by
$y=a{{\left( x-h \right)}^{2}}+k$
where “a” is the distance from the origin to the focus.
We are given that the parabola has vertex $\left( 2,-9 \right)$ . Here, we can see that $h=2$ and $k=-9$ . Hence, we can write the formula of parabola as
$y=a{{\left( x-2 \right)}^{2}}-9...\left( i \right)$
Now, we have to find a. We are given that the parabola passes through the point $\left( 12,-4 \right)$ . Let us substitute $x=12$ and $y=-4$ in the above equation.
$\Rightarrow -4=a{{\left( 12-2 \right)}^{2}}-9$
We have to solve for “a”. Let us first simplify the terms inside the bracket.
$\begin{align}
& \Rightarrow -4=a{{\left( 10 \right)}^{2}}-9 \\
& \Rightarrow -4=100a-9 \\
\end{align}$
Let us take -9 from RHS to the LHS.
$\begin{align}
& \Rightarrow -4+9=100a \\
& \Rightarrow 5=100a \\
& \Rightarrow 100a=5 \\
\end{align}$
We can now find a by taking 100 to the RHS.
$\Rightarrow a=\dfrac{5}{100}=\dfrac{1}{20}$
Now, we have to substitute this value of a in the equation (i).
$y=\dfrac{1}{20}{{\left( x-2 \right)}^{2}}-9$
Thus, the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through point $\left( 12,-4 \right)$ is $y=\dfrac{1}{20}{{\left( x-2 \right)}^{2}}-9$.
Note: Students must know the standard and vertex form of a parabola. Standard form is given by $y=a{{x}^{2}}+bx+c$ . We did not use this formula in this question because the standard formula does not involve vertex. Let us draw this parabola. We can see that a is positive. Therefore, the parabola opens upwards. If a is negative, then the parabola opens downwards.
Complete step-by-step solution:
We have to find the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through point $\left( 12,-4 \right)$ . We know that the equation of a parabola with vertex $\left( h,k \right)$ is given by
$y=a{{\left( x-h \right)}^{2}}+k$
where “a” is the distance from the origin to the focus.
We are given that the parabola has vertex $\left( 2,-9 \right)$ . Here, we can see that $h=2$ and $k=-9$ . Hence, we can write the formula of parabola as
$y=a{{\left( x-2 \right)}^{2}}-9...\left( i \right)$
Now, we have to find a. We are given that the parabola passes through the point $\left( 12,-4 \right)$ . Let us substitute $x=12$ and $y=-4$ in the above equation.
$\Rightarrow -4=a{{\left( 12-2 \right)}^{2}}-9$
We have to solve for “a”. Let us first simplify the terms inside the bracket.
$\begin{align}
& \Rightarrow -4=a{{\left( 10 \right)}^{2}}-9 \\
& \Rightarrow -4=100a-9 \\
\end{align}$
Let us take -9 from RHS to the LHS.
$\begin{align}
& \Rightarrow -4+9=100a \\
& \Rightarrow 5=100a \\
& \Rightarrow 100a=5 \\
\end{align}$
We can now find a by taking 100 to the RHS.
$\Rightarrow a=\dfrac{5}{100}=\dfrac{1}{20}$
Now, we have to substitute this value of a in the equation (i).
$y=\dfrac{1}{20}{{\left( x-2 \right)}^{2}}-9$
Thus, the equation of the parabola that has a vertex at $\left( 2,-9 \right)$ and passes through point $\left( 12,-4 \right)$ is $y=\dfrac{1}{20}{{\left( x-2 \right)}^{2}}-9$.
Note: Students must know the standard and vertex form of a parabola. Standard form is given by $y=a{{x}^{2}}+bx+c$ . We did not use this formula in this question because the standard formula does not involve vertex. Let us draw this parabola. We can see that a is positive. Therefore, the parabola opens upwards. If a is negative, then the parabola opens downwards.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

