Find the coordinates of a point on the parabola $y^2 = 8x$ whose focal distance is 4.

> [!TLDR] > For a parabola $y^2 = 4ax$, the focal distance of any point $(x, y)$ on it is $x+a$. By comparing $y^2 = 8x$ with the standard form, we find $a=2$. Setting the focal distance $x+a=4$ gives $x=2$, which leads to the points $\boxed{(2, \pm 4)}$. To find the coordinates of a point on a parabola given its focal distance, we first need to understand what focal distance means and how it relates to the standard form of a parabola. A parabola is defined as the locus of points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). This fundamental definition is key to solving this problem efficiently. For a parabola in the standard form $y^2 = 4ax$: * The **focus** is at the point $(a, 0)$. * The **directrix** is the vertical line $x = -a$. > [!FORMULA] > For a parabola $y^2 = 4ax$, the focal distance of any point $P(x_1, y_1)$ on the parabola is given by the distance from $P$ to the directrix $x=-a$. > Focal distance $= |x_1 - (-a)| = |x_1 + a|$. > Since $y^2 = 4ax$ implies $x \ge 0$ (as $y^2 \ge 0$ and $a > 0$ for $y^2=8x$), $x_1+a$ will always be positive. > Thus, Focal distance $= x_1 + a$. Now, let's apply this to the given problem. 1. **Identify the standard form and parameter 'a'.** The given equation of the parabola is $y^2 = 8x$. We compare this with the standard form $y^2 = 4ax$. By equating the coefficients of $x$, we get $4a = 8$. Solving for $a$, we find $a = \frac{8}{4} = 2$. 2. **Determine the focus and directrix.** With $a=2$: * The focus of the parabola is $(a, 0) = (2, 0)$. * The equation of the directrix is $x = -a$, which is $x = -2$. 3. **Use the focal distance property.** Let $P(x_1, y_1)$ be a point on the parabola whose focal distance is 4. According to the definition of a parabola, the focal distance of $P(x_1, y_1)$ is equal to its perpendicular distance from the directrix $x = -2$. The distance from a point $(x_1, y_1)$ to the line $x+2=0$ is $|x_1

Find the coordinates of a point on the parabola y² = 8x whose focal distance is 4.

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y^2 = 8x

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Concept understanding — Parabola Standard Form

Standard Equations of a Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). To turn this definition into a clean equation, NCERT places the parabola in the simplest position: vertex at the origin with its axis along a coordinate axis. The equations you get are called the standard equations.

The four standard forms

Depending on which way the parabola opens, there are four standard equations. In each, $a > 0$ .

y^2 = 4ax

Equation	Opens	Focus	Directrix
$y^2 = 4ax$	right	$(a, 0)$	$x = -a$
$y^2 = -4ax$	left	$(-a, 0)$	$x = a$
$x^2 = 4ay$	up	$(0, a)$	$y = -a$
$x^2 = -4ay$	down	$(0, -a)$	$y = a$

For all four the vertex is at the origin $(0,0)$ and the axis of the parabola is a coordinate axis.

Where $y^2 = 4ax$ comes from

Take the focus at $F(a, 0)$ and the directrix as the line $x = -a$ . For a point $P(x, y)$ on the parabola, its distance to the focus equals its distance to the directrix:

\sqrt{(x - a)^2 + y^2} = x + a.

Squaring both sides:

(x - a)^2 + y^2 = (x + a)^2,

and expanding gives $y^2 = 4ax$ . The other three forms follow by turning the focus in a different direction.

Latus rectum

The latus rectum is the chord through the focus, perpendicular to the axis, with both ends on the parabola. For every standard parabola its length is $4a$ — the very same $4a$ that appears in the equation, which makes it quick to read off.

Worked example

For the parabola $y^2 = 12x$ , compare with $y^2 = 4ax$ : here $4a = 12$ , so $a = 3$ .

Vertex: $(0, 0)$
Focus: $(a, 0) = (3, 0)$
Directrix: $x = -3$ …

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For a parabola $y^2 = 4ax$ , the focal distance of any point $(x, y)$ on it is $x+a$ . By comparing $y^2 = 8x$ with the standard form, we find $a=2$ . Setting the focal distance $x+a=4$ gives $x=2$ , which leads to the points $\boxed{(2, \pm 4)}$ .

To find the coordinates of a point on a parabola given its focal distance, we first need to understand what focal distance means and how it relates to the standard form of a parabola.

A parabola is defined as the locus of points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). This fundamental definition is key to solving this problem efficiently.

For a parabola in the standard form $y^2 = 4ax$ :

The focus is at the point $(a, 0)$ .
The directrix is the vertical line $x = -a$ .

y^2 = 4ax

Now, let's apply this to the given problem.

Identify the standard form and parameter 'a'.

The given equation of the parabola is $y^2 = 8x$ .

We compare this with the standard form $y^2 = 4ax$ .

By equating the coefficients of $x$ , we get $4a = 8$ .

Solving for $a$ , we find $a = \frac{8}{4} = 2$ .
Determine the focus and directrix.

With $a=2$ :
- The focus of the parabola is $(a, 0) = (2, 0)$ .
- The equation of the directrix is $x = -a$ , which is $x = -2$ .
Use the focal distance property.

Let $P(x_1, y_1)$ be a point on the parabola whose focal distance is 4.

According to the definition of a parabola, the focal distance of $P(x_1, y_1)$ is equal to its perpendicular distance from the directrix $x = -2$ .

The distance from a point $(x_1, y_1)$ to the line $x+2=0$ is $|x_1+2|$ .

Since $y^2 = 8x$ implies $x \ge 0$ , and $a=2$ is positive, $x_1+2$ will always be positive. …

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Q.Find the coordinates of a point on the parabola y2=8xy^2 = 8xy2=8x whose focal distance is 4.

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Q.Find the coordinates of a point on the parabola $y^2 = 8x$ whose focal distance is 4.

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