Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively

Q: Wait, why is the vertex $(2, 0)$ and not $(0, 2)$?

Great question! The problem states the vertex is at a distance of $2$ from the origin *on the positive x-axis*. This means the x-coordinate is $2$ and the y-coordinate is $0$. If it were on the y-axis, it would be $(0, 2)$. Always read the 'on the x-axis' part carefully!

Q: How do we know the parabola opens to the right and not the left?

Look at the positions: the vertex is at $x = 2$ and the focus is at $x = 4$. Since the focus is always inside the parabola, the curve must open towards the focus. Because $4 > 2$, it must open towards the positive x-direction (to the right).

Q: Can we use the general equation $y^2 + Dx + Ey + F = 0$ instead?

You absolutely can! However, using the standard form $(y - k)^2 = 4a(x - h)$ is much faster and less prone to algebraic errors when you already know the vertex and the axis. It directly maps to the geometric properties of the parabola.

Question

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

Options:

(A) $(4, -4)$

(B) $(5, 2\sqrt{6})$

(C) $(8, 6)$

(D) $(6, 4\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Parabola

Imagine you are standing on the Cartesian plane, looking at the $x$-axis. You are told that a parabola's axis of symmetry lies perfectly along this line.

This is a powerful piece of information. It tells us that for every point $(x, y)$ on the parabola, there is a corresponding point $(x, -y)$ on the other side. The parabola is balanced, like a mirror image across the $x$-axis.

### Locating the Heart of the Curve

The problem provides two critical landmarks: the vertex and the focus. The vertex is the turning point, the very tip of the parabola.

We are told it sits at a distance of $2$ units from the origin on the positive $x$-axis. Thus, our vertex $V$ is at $(2, 0)$.

The focus $F$ is the 'heart' of the parabola, the point that defines its curvature. It sits at a distance of $4$ units from the origin on the same axis, so $F$ is at $(4, 0)$.

Notice how the focus is to the right of the vertex. This confirms our parabola opens to the right, stretching out towards infinity in the positive $x$-direction.

### Defining the Focal Length

The distance between the vertex and the focus is denoted by $a$. This value $a$ is the secret key to the parabola's width.

Since the vertex is at $x = 2$ and the focus is at $x = 4$, the distance $a$ is simply $4 - 2 = 2$. This $a = 2$ tells us exactly how 'open' our parabola is.

### Constructing the Equation

Now, let's assemble our masterpiece. The standard equation for a parabola opening horizontally is $(y - k)^2 = 4a(x - h)$.

Here, $(h, k)$ are the coordinates of the vertex. Substituting our values, $h = 2$, $k = 0$, and $a = 2$, we get:

$$ (y - 0)^2 = 4(2)(x - 2) $$

Simplifying this, we arrive at the elegant identity of our curve:

$$ y^2 = 8(x - 2) $$

This equation is the rule that every point on our parabola must obey. If a point $(x, y)$ satisfies this equation, it belongs to the parabola.

### The Final Verification

Now, we test our options to find the one that does not belong.

For $(4, -4)$:
$y^2 = (-4)^2 = 16$, and $8(4 - 2) = 16$. It fits!

For $(5, 2\sqrt{6})$:
$y^2 = (2\sqrt{6})^2 = 4 	imes 6 = 24$, and $8(5 - 2) = 8 	imes 3 = 24$. It fits!

For $(8, 6)$:
$y^2 = 6^2 = 36$, but $8(8 - 2) = 8 	imes 6 = 48$. Since $36 
eq 48$, this point is not on the parabola.

For $(6, 4\sqrt{2})$:
$y^2 = (4\sqrt{2})^2 = 16 	imes 2 = 32$, and $8(6 - 2) = 8 	imes 4 = 32$. It fits!

We have successfully identified the outlier. The point **$(8, 6)$** is the only one that fails the test. Remember, in coordinate geometry, always build your equation first—it is your most reliable map.

Animated Solution for Mathematics - Conic Sections: Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin on the positive x-axis then which of the following points does not lie on it ?

Select Answer:

Visualized Solution

Visualizing the Coordinate System

Locating the Vertex $V (h, k)$

Locating the Focus $F$

Calculating Focal Length $a$

Standard Equation Form

Raw Setup (Substitution)

Atomic Compute (Execution)

Testing Point $(4, - 4)$

Testing Point $(5, 26)$

Testing Point $(8, 6)$

Testing Point $(6, 42)$

The Way Forward

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Geometry of the Parabola

Locating the Heart of the Curve

Defining the Focal Length

Constructing the Equation

The Final Verification

Similar Questions

If the vertices of a hyperbola be at $(- 2, 0)$ and $(2, 0)$ and one of its foci be at $(- 3, 0)$ , then which one of the following points does not lie on this hyperbola?

A hyperbola having the transverse axis of length $2$ has the same foci as that of the ellipse of $3 x^{2} + 4 y^{2} = 12$ , then this hyperbola does not pass through which of the following points?

The axis of a parabola is the line $y = x$ and its vertex and focus are in the first quadrant at distances $2$ and $22$ units from the origin, respectively. If the point $(1, k)$ lies on the parabola, then a possible value of $k$ is :-

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $2$ and $22$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

A parabola has the origin as its focus and the line $x = 2$ as the directrix. Then the vertex of the parabola is at

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

An ellipse, with foci at (0, 2) and (0, -2) and minor axis of length 4, passes through which of the following points ?

Consider a circle with its centre lying on the focus of the parabola $y^{2} = 2 p x$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and parabola is

Let $P$ and $Q$ be distinct points on the parabola $y^{2} = 2 x$ such that a circle with $P Q$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $Δ O P Q$ is $32$ , then which of the following is (are) the coordinates of $P$ ?