Question

Given the axis of a parabola lies along the x-axis. If its vertex and focus are at a distance 2 and 4 respectively from the origin on the positive x-axis, then which of the following points does not line on the parabola?
(a) $(4,-4)$,
(b) $(5,2\sqrt{6})$,
(c) $(8,6)$,
(d) $(6,4\sqrt{2})$.

Answer 1

Hint: We start solving the problem by finding the equation of the axis of parabola. We use the fact that vertex and focus lie on the axis of parabola to find and calculate them. We find the distance between vertices and focus to find the equation of the parabola. Once we find the equation of parabola, we substitute the points present in options to get the required answer.

Complete step-by-step answer:
According to the problem, the axis of the parabola lies along the x-axis and its vertex and focus area at a distance of 2 and 4 units respectively from the origin on the positive axis. We need to find which of the points given in options are not on the parabola.
We know that the equation of x-axis is $y=0$ and the axis of parabola is along x-axis. So, we get the axis of the parabola as $y=0$. We know that the vertex and focus of the parabola lies on the axis of the parabola. This makes the y-coordinate of focus and vertex as zero.
Since, the vertex of the parabola is 2 units from the origin in the positive direction of the x-axis and we know that the vertex lies on the axis of the parabola. So, we get the x-co-ordinate of the vertex as 2.
We get the vertex at the point $A(2,0)$.
Since, the focus of the parabola is 4 units from the origin in the positive direction of x-axis and we know that the focus lies on the axis of the parabola. So, we get the x-co-ordinate of the focus as 4.
We get the vertex at the focus $$B(4,0)$$.
Let us find the distance between the vertex and focus of the parabola. We know that distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$.
The distance between Vertex and focus is $AB=\sqrt{{(4-2)}^2+{(0-0)}^2}$.
$\Rightarrow AB=\sqrt{{\left(2\right)}^2+{\left(0\right)}^2}$.
$\Rightarrow AB=\sqrt{4}$.
$\Rightarrow AB=2$.
We know that equation of the parabola whose axis is $y=0$, having vertex at $(x_1,0)$ and distance between the focus and vertex as $a$ is $y^2=4a(x-x_1)$.
So, we get the equation of the parabola as $$y^2=(4\times 2)\times (x-2)$$.
$$\Rightarrow y^2=8\times (x-2)$$ ---(1).

Let us substitute $(4,-4)$ in equation (1) to verify whether the point is on parabola.
$$\Rightarrow {(-4)}^2=8\times (4-2)$$.
$$\Rightarrow 16=8\times (4-2)$$.
$$\Rightarrow 16=16$$.
The point $(4,-4)$ lies on the parabola.
Let us substitute $(5,2\sqrt{6})$ in equation (1) to verify whether the point is on parabola.
$$\Rightarrow {\left(2\sqrt{6}\right)}^2=8\times (5-2)$$.
$$\Rightarrow 4\times 6=8\times \left(3\right)$$.
$$\Rightarrow 24=24$$.
The point $(5,2\sqrt{6})$ lies on the parabola.
Let us substitute $(8,6)$ in equation (1) to verify whether the point is on parabola.
$$\Rightarrow {\left(6\right)}^2=8\times (8-2)$$.
$$\Rightarrow 36=8\times \left(6\right)$$.
$$\Rightarrow 36=48$$, which is a contradiction.
The point $(8,6)$ doesn’t lie on the parabola.
Let us substitute $(6,4\sqrt{2})$ in equation (1) to verify whether the point is on parabola.
$$\Rightarrow {\left(4\sqrt{2}\right)}^2=8\times (6-2)$$.
$$\Rightarrow 16\times 2=8\times \left(4\right)$$.
$$\Rightarrow 32=32$$.
The point $(6,4\sqrt{2})$ lies on the parabola.
We have found that the point $(8,6)$ does not lie on parabola.

So, the correct answer is “Option C”.

Note: We should not confuse the equation of x-axis with x=0 which leads to the wrong solution. We should know that the distance between focus and vertex is 4 times the length of the latus rectum of parabola. We should know that the parabola is symmetric about its axis. Similarly, we can expect problems to find the equation of tangent at a given point after finding the equation.