Position of a Point with Respect to a Parabola

The parabola drawn on the plane divides the plane into two regions, one is a concave (interior) region where the focus lies and the other is a convex (exterior) region. The chord of the parabola that passes through its focus is called the focal chord. Focus divides the focal chord into two segments which are called segments of the focal chord. In real life, we use Parabola as Parabolic reflectors like in telescopes.

This Story also Contains

1. Position of a Point wrt a Parabola
2. How to Find the Position of a Point with Respect to a Parabola
3. Solved Examples Based on the Position of a Point and Focal Chord

In this article, we will cover the concept of the Position of a Point with Respect to a Parabola. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of twenty questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2020, and one in 2021.

Position of a Point wrt a Parabola

The parabola drawn on the plane divides the plane into two regions, one is a concave (interior) region where the focus lies and the other is a convex (exterior) region.

How to Find the Position of a Point with Respect to a Parabola

The point can lie at three positions in a parabola.

a) Inside the parabola: The region in which the focus of a parabola is included constitutes the inside of the parabola. Any point present here is called the interior point.

b) On the parabola: The boundary of the parabola is called the region on the parabola.

c) Outside the parabola: The remaining region in which the focus is absent is called the outside of the parabola. Any point present here is called the exterior point.

Consider the parabola having equation $y^2=4 a x$
Let $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be a point in the plane (in the first or fourth quadrant). From point P , draw a perpendicular to the x -axis meeting the x -axis at M and a parabola at Q .

Let the coordinates of Q be $\left(\mathrm{x}_1, \mathrm{y}_2\right)$.
So, $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ any point in the plane and point $\mathrm{Q}\left(\mathrm{x}_1, \mathrm{y}_2\right)$ is on the parabola

The point P lies outside, on, and inside of the parabola according as

$
\begin{array}{ll}
& P M>,=\text { or }<Q M \\
\Rightarrow & P M^2>,=\text { or }<Q M^2 \\
\Rightarrow & y_1^2>,=\text { or }<y_2^2 \\
\Rightarrow & y_1^2>,=\text { or }<4 a x_1
\end{array}
$
Condition to find the position of Point
(a) P lies outside the parabola, $\mathrm{y}_1^2-4 \mathrm{ax}_1>0$
(b) P lies on the parabola, $y_1^2-4 \mathrm{ax}_1=0$
(c) P lies inside the parabola, $\mathrm{y}_1^2-4 \mathrm{ax}_1<0$

If point P lies in the second or third quadrant, $y_1^2-4 a x_1>0$ (as $\left.x_1<0\right)$.
Thus, $y^2-4 a x<0$ represents the concave region, and $y^2-4 a x>0$ represents the convex region of the parabola.

Similarly, $y^2+4 a x<0, x^2-4 a y<0$ and $x^2+4 a y<0$ represents the concave (interior) region of the parabolas.

Whereas, $y^2+4 a x>0, x^2-4 a y>0$ and $x^2+4 a y>0$ represent the convex(exterior) region of the parabolas $(a>0)$.

Recommended Video Based on the Position of a Point and Focal Chord

Solved Examples Based on the Position of a Point and Focal Chord

Example 1: Let $y=m x+c, m>0$ be the focal chord of $y^2=-64 x$ which is tangent to $(x+10)^2+y^2=4$. Then, the value of $4 \sqrt{2}(\mathrm{~m}+\mathrm{c})$ is equal to . $\qquad$
[JEE MAINS 2021]
Solution: Focus of $y^2=-64 x \Rightarrow y^2=-4(16) x$ is $(-16,0)$.
As $y=m x+c$ is the focal chord to $y^2=-64 x$

$
\begin{aligned}
& \Rightarrow \text { It passes through its focus }(-16,0) \\
& \Rightarrow \quad 0=m(-16)+c \Rightarrow c=16 m \\
& \therefore \text { Line is } y=m x+16 m
\end{aligned}
$
Now, this line is tangent to the circle

$
(x+10)^2+y^2=4 \quad(\text { centre }:(-10,0), \text { radius }=2)
$
So, the distance between the center of the circle $\&$ line $=$ radius of the circle

$
\begin{aligned}
& \Rightarrow \frac{|0+10 m-16 m|}{\sqrt{1+m^2}}=2 \\
& \Rightarrow|-6 m|=2 \sqrt{1+m^2} \\
& \Rightarrow 36 m^2=4+4 m^2 \Rightarrow m^2=\frac{4}{32}=\frac{1}{8} \\
& \Rightarrow m=\frac{1}{2 \sqrt{2}} \quad(\text { given } m>0)
\end{aligned}
$

$\& c=16 m=\frac{8}{\sqrt{2}}$
So $4 \sqrt{2}(m+c)=4 \sqrt{2}\left(\frac{1}{2 \sqrt{2}}+\frac{8}{\sqrt{2}}\right)=34$

Hence, the answer is 34

Example 2: The area of an equilateral triangle inscribed in the parabola $y^2=8 x$ with one of its vertices on the vertex of this parabola is:
[JEE MAINS 2020]
Solution

Let the two vertices of the triangle be $Q$ and $R$. Points $Q$ and $R$ will have the same $X$-coordinate $=k($ say )

Now in the right triangle PRT, right-angled at T.

$
\tan 30^{\circ}=\frac{\mathrm{RT}}{\mathrm{k}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{RT}}{\mathrm{k}} \Rightarrow \mathrm{RT}=\frac{\mathrm{k}}{\sqrt{3}} \Rightarrow \mathrm{R}\left(\mathrm{k}, \frac{\mathrm{k}}{\sqrt{3}}\right)
$
Now R lies on the parabola: $y^2=4 a x$

$
\begin{aligned}
& \Rightarrow\left(\frac{k}{\sqrt{3}}\right)^2=4 a(k) \\
& \Rightarrow \frac{k}{3}=4 a \\
& \Rightarrow k=12 a
\end{aligned}
$

Length of side of the triangle $=2(R T)=2 \cdot \frac{\mathrm{K}}{\sqrt{3}}=2 \cdot \frac{(12 \mathrm{a})}{\sqrt{3}}=8 \sqrt{3} \mathrm{a}=16 \sqrt{3}$
Area of Equilateral triangle is $=\frac{\sqrt{3}}{4} a^2=192 \sqrt{3}$

Hence, the answer is $192 \sqrt{3}$.

Example 3: The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is:
Solution: We know that the perpendicular distance of a point from a line - $\rho=\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$ $\rho$ is the distance from the line $a x+b y+c=0$.

Let $\left(a^2, a\right)$ be a point on $x=y^2$
Distance between $\left(a^2, a\right)$ and $x-y+1=$ is

$
\frac{a^2-a+1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left[\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right]
$
It is minimal when $a=\frac{1}{2}$
So minimum distance $=\frac{3}{4 \sqrt{2}}=\frac{3 \sqrt{2}}{8}$

Hence, the answer is the $\frac{3 \sqrt{2}}{8}$.

Example 4: A circle cuts a chord of length 4a on the X -axis and passes through a point on the Y axis, distant $2 b$ from the origin. Then, the locus of the center of this circle, is

Solution: According to the given information, we have the following figure

Let the equation of the circle be

$
x^2+y^2+2 g x+2 f y+c=0
$
According to the problem,

$
4 a=2 \sqrt{g^2-c}
$

$\qquad$
$\left[\because\right.$ The length of intercepts made by the circle $x^2+y^2+2 g x+2 f y+c=0$ with X -axis is $2 \sqrt{g^2-c}$

Also, as the circle is passing through

$
\begin{aligned}
& \therefore o+4 b^2+0+4 b f+c=0 \\
\Rightarrow & 4 b^2+4 b f+c=0
\end{aligned}
$
So, the locus is

$
\begin{aligned}
& 4 b^2-4 b f+x^2-4 a^2=0 \\
\Rightarrow & x^2=4 b y+4 a^2-4 b^2
\end{aligned}
$

which is a parabola.

Example 5: Let P be the point on the parabola $y^2=8 x$ which is at a minimum distance from the center C of the circle $x^2+\left(y+6^2\right)=1$. Then, the equation of the circle passing through $C$ and having its center at $P$ is
Solution: The Centre of the circle $x^2+(y+6)^2$ is $C(0,-6)$
Let the coordinates of point P be $\left(2 t^2, 4 t\right)$
Now, let

$
\begin{aligned}
& D=C P \\
\Rightarrow & D=\sqrt{\left(2 t^2\right)^2+(4 t+6)^2} \\
\Rightarrow & D=\sqrt{4 t^4+16 t^2+36+48 t} \\
\Rightarrow & D^2=4 t^4+16 t^2+36+48 t \\
\text { Let } \Rightarrow & F(t)=4 t^4+16 t^2+36+48 t
\end{aligned}
$
For minimum,

$
\begin{aligned}
& F^{\prime}(t)=0 \\
\Rightarrow & 16 t^3+32 t+48=0 \\
\Rightarrow & t^3+2 t+3=0 \\
\Rightarrow & (t+1)\left(t^2-t+3\right)=0 \\
\Rightarrow & t=-1
\end{aligned}
$
Thus, the coordinates of point P are $(2,-4)$

$
C P=\sqrt{2^2+(-4+6)^2}=\sqrt{4+4}=1 \sqrt{2}
$
Hence, the required equation for a circle is

$\begin{aligned} & (x-2)^2+(y+4)^2=(2 \sqrt{2})^2 \\ & \Rightarrow x^2+y^2-4 x+8 y+12=0\end{aligned}$

Hence, the answer is $x^2+y^2-4 x+8 y+12=0$