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Question: If the vertex of the parabola is \[\left( 2,0 \right)\] and the extremities of the latus rectum are ...

If the vertex of the parabola is (2,0)\left( 2,0 \right) and the extremities of the latus rectum are (3,2)\left( 3,2 \right)and (3,−2)\left( 3,-2 \right) then the equation of the parabola is

& \text{A}\text{. }{{\text{y}}^{2}}=2x-4 \\\ & \text{B}\text{. }{{\text{x}}^{2}}=4y-8 \\\ & \text{C}\text{. }{{\text{y}}^{2}}=4x-8 \\\ & \text{D}\text{. None of these} \\\ \end{aligned}$$
Explanation

Solution

We know that if the latus rectum is perpendicular to y-axis and the vertex of parabola is (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right), then the equation of the parabola is (y−y1)2=4a(x−x1){{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right). In the same way, if the latus rectum is perpendicular to x-axis and the vertex of parabola is (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right), then the equation of the parabola is (x−x1)2=4a(y−y1){{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right). From the question, it is given that the vertex of the parabola is (2,0)\left( 2,0 \right) and the extremities of the latus rectum are (3,2)\left( 3,2 \right)and (3,−2)\left( 3,-2 \right). So, we will find the equation of latus rectum. Then by using the above concept, we will find the equation of parabola.

Complete step-by-step answer:
Before solving the question, we should know that if the latus rectum is perpendicular to y-axis and the vertex of parabola is (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right), then the equation of the parabola is (y−y1)2=4a(x−x1){{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right). In the same way, if the latus rectum is perpendicular to x-axis and the vertex of parabola is (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right), then the equation of the parabola is (x−x1)2=4a(y−y1){{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right).

From the question, it is given that the vertex of the parabola is (2,0)\left( 2,0 \right) and the extremities of the latus rectum are (3,2)\left( 3,2 \right)and (3,−2)\left( 3,-2 \right).

Now we have to find the equation of the latus rectum.
We know that if A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) are two points, then the line passing through A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is y−y1=(y2−y1x2−x1)(x−x1)y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right).
Now we have to find the equation of line passing through (3,2)\left( 3,2 \right)and (3,−2)\left( 3,-2 \right).

& y-2=\dfrac{-2-2}{3-3}\left( x-3 \right) \\\ & \Rightarrow y-2=\dfrac{-4}{0}\left( x-3 \right) \\\ & \Rightarrow x=3 \\\ \end{aligned}$$ We know that a line $$x=k$$ is perpendicular to the y-axis. Hence, we can say that the latus rectum is perpendicular to y-axis. We should know that if the latus rectum is perpendicular to y-axis and the vertex of parabola is $$\left( {{x}_{1}},{{y}_{1}} \right)$$, then the equation of the parabola is $${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$$. So, the equation of parabola whose vertex is $$\left( 2,0 \right)$$ is $$\begin{aligned} & {{\left( y-0 \right)}^{2}}=4a\left( x-2 \right) \\\ & \Rightarrow {{y}^{2}}=4a\left( x-2 \right).....(1) \\\ \end{aligned}$$ We know that the extremities of the latus rectum passes through the parabola. So, we can say that $$\left( 3,2 \right)$$and $$\left( 3,-2 \right)$$ lie on equation (1). So, let us substitute $$\left( 3,2 \right)$$ in equation (1). $$\begin{aligned} & \Rightarrow {{\left( 2 \right)}^{2}}=4a\left( 3-2 \right) \\\ & \Rightarrow 4=4\left( a \right) \\\ & \Rightarrow a=1....(2) \\\ \end{aligned}$$ Now let us substitute equation (2) in equation (1), then we get $$\begin{aligned} & \Rightarrow {{y}^{2}}=4\left( x-2 \right) \\\ & \Rightarrow {{y}^{2}}=4x-8 \\\ \end{aligned}$$ **So, the correct answer is “Option C”.** **Note:** Students may have a misconception that if the latus rectum is perpendicular to x-axis and the vertex of parabola is $$\left( {{x}_{1}},{{y}_{1}} \right)$$, then the equation of the parabola is $${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$$. In the same way, if the latus rectum is perpendicular to y-axis and the vertex of parabola is $$\left( {{x}_{1}},{{y}_{1}} \right)$$, then the equation of the parabola is $${{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right)$$. To avoid this conception, students should draw the graph of the parabola. By observing the graph, students may have a clear view.