How do we represent a chord of a parabola passing through the origin?

Take a line through origin, e.g. $y=mx$. Its intersection points with the parabola give a chord through the origin. This is the standard parameter method for such questions.

What is the equation of parabola y^2 = 4ax in standard form?

The standard right-opening parabola is $y^2=4ax$, with vertex at origin and focus $(a,0)$. Here $a=1$, so $y^2=4x$.

How to find the other intersection point of y = mx with y^2 = 4x?

Substitute $y=mx$ into $y^2=4x$ to get $m^2x^2=4x$. One solution is $x=0$ (origin) and the other is $x=\frac{4}{m^2}$, giving $y=\frac{4}{m}$.

How to get the locus of mid-point of a variable chord?

Find midpoint coordinates in terms of parameter (here slope $m$), then eliminate the parameter to obtain an equation only in $x$ and $y$. That equation is the locus.

What is the section formula for internal division in ratio m:n?

If a point divides segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in ratio $m:n$, then coordinates are $\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$ when taken as $A$ to $B$.

If O is the origin, how does the section formula simplify?

If $O(0,0)$ and $P(x_1,y_1)$, then the point dividing $OP$ internally in ratio $m:n$ (i.e. $OR:RP=m:n$) is $\left(\frac{m}{m+n}x_1,\frac{m}{m+n}y_1\right)$.

Why does scaling a locus work here?

The dividing point on $OP$ is a constant fraction of $P$, so its coordinates are scaled versions of $P$'s coordinates. Replacing $x_1,y_1$ by multiples of $x,y$ in the locus equation yields the new locus.

What is the locus S of midpoints in this problem?

For chords through origin of $y^2=4x$, the midpoint locus becomes $y^2=2x$. This is found by parameter elimination using line $y=mx$.

What is the final locus after dividing OP in ratio 3:1?

If $P$ lies on $y^2=2x$ and $R$ divides $OP$ in ratio $3:1$, then $R(\frac{3}{4}x_1,\frac{3}{4}y_1)$. Eliminating $x_1,y_1$ gives the locus $2y^2=3x$.

Let the locus of the mid-point of the chord through the origin O of the parabola y^2 = 4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 : 1, is :

Let the locus of the mid-point of the chord through the origin O of the parabola y^2 = 4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 : 1, is : | JEE Main Mathematics

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Math Locus

Q MCQ Parabola

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

(A) $2x^2=3y$
(B) $2y^2=3x$
(C) $3y^2=2x$
(D) $3x^2=2y$

✅ Correct Answer

Option B

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Solution Steps

Take a general chord through the origin

Any chord through the origin $O(0,0)$ can be obtained by taking a line through the origin:

$$y=mx \quad (m\ne 0).$$

This line intersects the parabola $y^2=4x$ at $O$ and one more point.

Find the second intersection point with the parabola

Substitute $y=mx$ in $y^2=4x$:

$$m^2x^2=4x \;\Rightarrow\; x(m^2x-4)=0.$$

So $x=0$ gives the origin, and the other point has

$$x=\frac{4}{m^2},\qquad y=mx=\frac{4}{m}.$$

Let this point be $A\left(\frac{4}{m^2},\frac{4}{m}\right)$.

Write the midpoint of chord OA and get curve S

Midpoint $M$ of $O(0,0)$ and $A\left(\frac{4}{m^2},\frac{4}{m}\right)$ is:

$$M\left(\frac{2}{m^2},\frac{2}{m}\right).$$

Let $M(x,y)$ belong to locus $S$. Then:

$$x=\frac{2}{m^2},\quad y=\frac{2}{m}\Rightarrow m=\frac{2}{y}.$$

Substitute in $x=\frac{2}{m^2}$:

$$x=\frac{2}{(2/y)^2}=\frac{y^2}{2}\;\Rightarrow\; y^2=2x.$$

Hence curve $S$ is $y^2=2x$.

Let P be any point on S

Since $P$ lies on $S$, write:

$$P(x_1,y_1)\ \text{with}\ y_1^2=2x_1.$$

Now we need the locus of point $R$ which divides $OP$ internally in ratio $3:1$.

Use section formula for division of OP in ratio 3:1

Origin is $O(0,0)$. If $R$ divides $OP$ internally as $OR:RP=3:1$, then:

$$R\left(\frac{3x_1+1\cdot 0}{4},\frac{3y_1+1\cdot 0}{4}\right)=\left(\frac{3x_1}{4},\frac{3y_1}{4}\right).$$

Let $R(x,y)$. Then:

$$x=\frac{3x_1}{4},\quad y=\frac{3y_1}{4}\Rightarrow x_1=\frac{4x}{3},\ y_1=\frac{4y}{3}.$$

Eliminate P using y₁²=2x₁

Use $y_1^2=2x_1$ with $x_1=\frac{4x}{3}$ and $y_1=\frac{4y}{3}$:

$$\left(\frac{4y}{3}\right)^2=2\left(\frac{4x}{3}\right).$$

$$\frac{16y^2}{9}=\frac{8x}{3}\;\Rightarrow\;16y^2=24x\;\Rightarrow\;2y^2=3x.$$

This is exactly option (B).

Conclusion

The required locus is:

$$\boxed{2y^2=3x}$$

So the correct answer is $\boxed{\text{Option (B)}}$.

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Key Insight

Chord through origin ⇒ use $y=mx$. Midpoints give $S: y^2=2x$. Ratio point scales coordinates by $\frac{3}{4}$ ⇒ locus becomes $2y^2=3x$.

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Coordinate Geometry Theory

1. Chord through the origin for a parabola: the slope-parameter method

In locus questions involving a parabola and a chord passing through a fixed point, the fastest approach is to use a one-parameter family of lines through that fixed point. Here the chord must pass through the origin, so every such line can be written as $y=mx$. When you substitute this line into the parabola $y^2=4x$, you get the intersection points in terms of the parameter $m$. One of those points is automatically the origin, and the other point gives the second end of the chord. This method is powerful because it converts a “moving chord” into a single algebraic parameter. In JEE problems, it is usually better than trying to write the general chord equation directly, because the general chord form can be messy. Once the second intersection point is found, midpoint, section point, or any other derived point can be computed using coordinate formulas, and then the parameter can be eliminated to obtain the locus.

2. Locus of a midpoint: compute in terms of parameter and eliminate

A midpoint locus problem almost always follows a fixed routine: (i) express the endpoints of the segment in terms of one parameter, (ii) compute midpoint coordinates using the midpoint formula, and (iii) eliminate the parameter to get an equation in $x$ and $y$ only. In this question, the chord endpoints are $O(0,0)$ and $A\left(\frac{4}{m^2},\frac{4}{m}\right)$, so the midpoint is $M\left(\frac{2}{m^2},\frac{2}{m}\right)$. Now you set $M(x,y)$ and remove $m$ by using $y=\frac{2}{m}\Rightarrow m=\frac{2}{y}$, then substitute into $x=\frac{2}{m^2}$. This elimination step is where the locus appears. Notice that the resulting locus $y^2=2x$ is itself a parabola. Such “midpoint loci” often produce a similar curve but scaled or shifted; recognizing this helps you verify your final equation quickly.

3. Internal division of a segment: section formula and ratio points

The section formula is a key coordinate-geometry tool used in JEE locus questions. If a point divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m:n$ (i.e., $AP:PB=m:n$), then the dividing point is $\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$. A special simplification happens when one endpoint is the origin. If $O(0,0)$ and $P(x_1,y_1)$, then the point $R$ dividing $OP$ in ratio $m:n$ satisfies $R\left(\frac{m}{m+n}x_1,\frac{m}{m+n}y_1\right)$. This means the ratio point is just a scaling of $P$ by a constant factor. In the given problem, the factor is $\frac{3}{4}$, so the coordinates shrink uniformly. This directly leads to a new locus obtained by rescaling variables in the equation of curve $S$.

4. Scaling a locus: how equations transform under coordinate scaling

Many JEE locus questions can be finished quickly if you understand how a curve changes when points are scaled from the origin. Suppose a point $P(x_1,y_1)$ lies on a curve and a new point $R$ is defined by $R(kx_1,ky_1)$ for a constant $k$ (with $0

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FAQs

What line represents a chord through the origin?

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Any line through origin can be written as $y=mx$. Intersecting it with the parabola gives the chord endpoints.

How to find the second intersection with $y^2=4x$?

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Substitute $y=mx$ to get $m^2x^2=4x$. Besides $x=0$, the other solution is $x=4/m^2$ and $y=4/m$.

What is the midpoint of $O(0,0)$ and $\left(\frac{4}{m^2},\frac{4}{m}\right)$?

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It is $M\left(\frac{2}{m^2},\frac{2}{m}\right)$ by the midpoint formula.

What is the locus S of these midpoints?

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Eliminating $m$ gives $y^2=2x$. So curve $S$ is the parabola $y^2=2x$.

How to apply section formula when one point is the origin?

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If $R$ divides $OP$ in ratio $m:n$ with $O(0,0)$, then $R=\left(\frac{m}{m+n}x_1,\frac{m}{m+n}y_1\right)$.

What are the coordinates of the point dividing $OP$ in ratio $3:1$?

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If $P(x_1,y_1)$, then $R=\left(\frac{3x_1}{4},\frac{3y_1}{4}\right)$.

How do we eliminate $P$ to get the final locus?

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Use $x_1=\\frac{4x}{3}, y_1=\\frac{4y}{3}$ in $y_1^2=2x_1$ and simplify to get the locus.

Why does the locus remain a parabola?

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The mapping from $P$ to the ratio point is just a scaling from the origin, which preserves the parabola type (only changes coefficients).

What is the final locus equation?

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The final locus is $2y^2=3x$.

Which option is correct?

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Option (B) $2y^2=3x$ is correct.

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