JEE Main 2026 (22 January Shift 2)

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Animated Solution for Mathematics - Conic Sections: Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2} = 4 x$ be the curve $S$ . Let $P$ be any point on $S$ . Then the locus of the point, which internally divides $O P$ in the ratio $3 : 1$ , is :

Q: Can we use the general equation of a chord passing through the origin?

Absolutely! You could assume the chord is $y = mx$. Substituting this into $y^2 = 4x$ gives $(mx)^2 = 4x$, which simplifies to $m^2x^2 - 4x = 0$. The roots are $x=0$ (the origin) and $x = \frac{4}{m^2}$. This gives you the intersection point $A$. While this method works, the parametric approach is often faster and less prone to division-by-zero errors if the chord is vertical.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the chord passing through the origin

O (0, 0)

and intersecting the parabola

y^{2} = 4 x

. If you represent the intersection point

A

using a parameter

t

, what are its coordinates? Remember, the midpoint

M

O A

will also depend on this parameter

t

Hint 2

The Arsenal: To find the locus of the midpoint

M (h, k)

, you need to express

h

and

k

in terms of

t

and then eliminate

t

. Once you have the equation for the curve

S

, treat

P (x_{1}, y_{1})

as a point on

S

. Use the section formula to find the coordinates of the point

R (h, k)

that divides

O P

in the ratio

3 : 1

Hint 3

The Execution: Start by setting

A = (t^{2}, 2 t)

. The midpoint

M (h, k)

(\frac{t ^{2}}{2}, t)

. This leads to

h = \frac{k ^{2}}{2}

, or

y^{2} = 2 x

for the curve

S

. Now, for a point

P (x_{1}, y_{1})

S

, the point

R (h, k)

dividing

O P

3 : 1

is given by

h = \frac{3 x _{1} + 1 ( 0 )}{4}

and

k = \frac{3 y _{1} + 1 ( 0 )}{4}

. Substitute

x_{1} = \frac{4 h}{3}

and

y_{1} = \frac{4 k}{3}

into the equation of

S

Visualized Solution

The Given Parabola

Given Parabola: $y^{2} = 4 x$
Origin is $O (0, 0)$ .
Let a chord pass through $O$ and intersect the parabola at $A$ .

Parametric Coordinates of $A$

Any point on $y^{2} = 4 a x$ can be taken as $(a t^{2}, 2 a t)$ .
Here, $4 a = 4 ⟹ a = 1$ .
So, point $A$ is $(t^{2}, 2 t)$ .

Mid-point of Chord $O A$

Let the mid-point of $O A$ be $M (h, k)$ .
Using the mid-point formula:
$h = \frac{t ^{2} + 0}{2} = \frac{t ^{2}}{2}$
$k = \frac{2 t + 0}{2} = t$

Finding the Curve $S$

We need to eliminate the parameter $t$ .
From $k = t$ , substitute $t$ into the equation for $h$ .
$h = \frac{k ^{2}}{2} ⟹ k^{2} = 2 h$
Replacing $(h, k)$ with $(x, y)$ , the locus $S$ is $y^{2} = 2 x$ .

Point $P$ on Curve $S$

Curve $S$ is the parabola $y^{2} = 2 x$ .
Let $P (x_{1}, y_{1})$ be any point on this curve $S$ .
Therefore, it must satisfy $y_{1}^{2} = 2 x_{1}$ .

Section Formula Setup

We need the locus of a point dividing $O P$ internally in the ratio $3 : 1$ .
Let this new point be $R (h, k)$ .
The line segment is from $O (0, 0)$ to $P (x_{1}, y_{1})$ .

Applying Section Formula

Using $m : n = 3 : 1$ :
$h = \frac{3 ( x _{1} ) + 1 ( 0 )}{3 + 1} = \frac{3 x _{1}}{4}$
$k = \frac{3 ( y _{1} ) + 1 ( 0 )}{3 + 1} = \frac{3 y _{1}}{4}$

Expressing $P$ in terms of $R$

From $h = \frac{3 x _{1}}{4}$ , we get $x_{1} = \frac{4 h}{3}$ .
From $k = \frac{3 y _{1}}{4}$ , we get $y_{1} = \frac{4 k}{3}$ .

Substituting into Curve $S$

We know $P (x_{1}, y_{1})$ lies on $S : y_{1}^{2} = 2 x_{1}$ .
Substitute $x_{1}$ and $y_{1}$ :
$(\frac{4 k}{3})^{2} = 2 (\frac{4 h}{3})$

Final Locus Equation

Expanding: $\frac{16 k ^{2}}{9} = \frac{8 h}{3}$
Simplifying: $\frac{2 k ^{2}}{3} = h ⟹ 2 k^{2} = 3 h$
Replacing $(h, k)$ with $(x, y)$ , the final locus is $2 y^{2} = 3 x$ .

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

Solution Diagram

Analyzing the Setup

We are exploring the geometry of a parabola defined by the equation

y^{2} = 4 x

. Our objective is to track the locus of a point undergoing a two-stage transformation.

First, we consider a chord originating from the origin

O (0, 0)

to a point

A

on the parabola. We aim to find the path traced by the midpoint

M

of this chord.

The Birth of Curve $S$

To simplify the geometry, we utilize parametric coordinates. For the parabola

y^{2} = 4 x

, where

a = 1

, any point

A

can be represented as

(t^{2}, 2 t)

The midpoint

M (h, k)

of the segment

O A

is calculated using the midpoint formula:

h = \frac{t ^{2} + 0}{2} = \frac{t ^{2}}{2}, k = \frac{2 t + 0}{2} = t

To find the locus of

M

, we eliminate the parameter

t

. Substituting

t = k

into the expression for

h

, we obtain

h = \frac{k ^{2}}{2}

, which simplifies to

k^{2} = 2 h

Replacing

(h, k)

with

(x, y)

, we identify the curve

S

as:

y^{2} = 2 x

The Transformation of $P$

Next, let

P (x_{1}, y_{1})

be any point on the curve

S

. Since

P

lies on

S

, it must satisfy the constraint:

y_{1}^{2} = 2 x_{1}

We now define a point

R (h, k)

that divides the segment

O P

internally in the ratio

3 : 1

. Applying the section formula, the coordinates of

R

are:

h = \frac{3 ( x _{1} ) + 1 ( 0 )}{3 + 1} = \frac{3 x _{1}}{4}, k = \frac{3 ( y _{1} ) + 1 ( 0 )}{3 + 1} = \frac{3 y _{1}}{4}

The Final Reveal

To determine the locus of

R

, we express the original coordinates

(x_{1}, y_{1})

in terms of

(h, k)

x_{1} = \frac{4 h}{3}, y_{1} = \frac{4 k}{3}

Substituting these expressions into the equation for curve

S

(

y_{1}^{2} = 2 x_{1}

), we get:

(\frac{4 k}{3})^{2} = 2 (\frac{4 h}{3})

Expanding the terms, we have:

\frac{16 k ^{2}}{9} = \frac{8 h}{3}

Multiplying both sides by

\frac{9}{8}

to isolate the variables, we obtain

2 k^{2} = 3 h

. Replacing

(h, k)

with

(x, y)

, the final locus is:

$2 y^{2} = 3 x$

Common Traps & FAQs

Why do we use parametric coordinates

(t^{2}, 2 t)

instead of just

(x, y)

Great question! Using parametric coordinates is a powerful tool in coordinate geometry. It reduces the number of variables we have to juggle. By expressing both

x

and

y

in terms of a single parameter

t

, we can easily eliminate it to find the locus. If we used

(x, y)

directly, we would be dealing with a more complex system of equations involving the chord's slope, which is much more prone to algebraic errors.

Wait, why do we replace

(h, k)

with

(x, y)

at the very end?

That is a standard convention in mathematics. When we define a point

(h, k)

as a variable point on a locus, we are essentially looking for a relationship between its coordinates. Once we find that relationship, say

2 k^{2} = 3 h

, we rename

(h, k)

(x, y)

to represent the general equation of the curve in the Cartesian plane. It is just a way of saying, 'Any point

(x, y)

on this curve must satisfy this equation.'

Can we use the general equation of a chord passing through the origin?

Absolutely! You could assume the chord is

y = m x

. Substituting this into

y^{2} = 4 x

gives

(m x)^{2} = 4 x

, which simplifies to

m^{2} x^{2} - 4 x = 0

. The roots are

x = 0

(the origin) and

x = \frac{4}{m ^{2}}

. This gives you the intersection point

A

. While this method works, the parametric approach is often faster and less prone to division-by-zero errors if the chord is vertical.

Animated Solution for Mathematics - Conic Sections: Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2} = 4 x$ be the curve $S$ . Let $P$ be any point on $S$ . Then the locus of the point, which internally divides $O P$ in the ratio $3 : 1$ , is :

Select Answer:

Visualized Solution

The Given Parabola

Parametric Coordinates of $A$

Mid-point of Chord $O A$

Finding the Curve $S$

Point $P$ on Curve $S$

Section Formula Setup

Applying Section Formula

Expressing $P$ in terms of $R$

Substituting into Curve $S$

Final Locus Equation

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

Analyzing the Setup

The Birth of Curve $S$

The Transformation of $P$

The Final Reveal

Similar Questions

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Select Answer:

Visualized Solution

The Given Parabola

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Mid-point of Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA

Finding the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Section Formula Setup

Applying Section Formula

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Substituting into Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Final Locus Equation

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

Analyzing the Setup

The Birth of Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Transformation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Final Reveal

Similar Questions

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Parametric Coordinates of $A$

Mid-point of Chord $O A$

Finding the Curve $S$

Point $P$ on Curve $S$

Expressing $P$ in terms of $R$

Substituting into Curve $S$

The Birth of Curve $S$

The Transformation of $P$

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then