Maths Q11 – JEE-MAIN 2026

Let O be the vertex of the parabola

y^{2} = 4 x

and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is :

Clue 1 of 3: Geometric Properties

The vertex of the parabola $y^{2} = 4 x$ is at the origin $(0, 0)$ . Use parametric coordinates for points on the parabola.

🧠 Full Solution Path

Step 1: Parametric Representation and Perpendicularity Condition✦ Active

The given parabola is $y^{2} = 4 x$ . Its vertex O is at $(0, 0)$ , and the parameter $a = 1$ . Let the points P and Q on the parabola be represented by their parametric coordinates as $P (t_{1}^{2}, 2 t_{1})$ and $Q (t_{2}^{2}, 2 t_{2})$ . The slopes of the chords OP and OQ are $m_{O P} = \frac{2 t_{1} - 0}{t_{1}^{2} - 0} = \frac{2}{t_{1}}$ and $m_{O Q} = \frac{2 t_{2} - 0}{t_{2}^{2} - 0} = \frac{2}{t_{2}}$ . Since OP and OQ are perpendicular, the product of their slopes is $- 1$ :

m_{O P} \cdot m_{O Q} = - 1 ⟹ \frac{2}{t_{1}} \cdot \frac{2}{t_{2}} = - 1 ⟹ t_{1} t_{2} = - 4

Step 2: Determine the Locus of the Mid-point○ Expand

Let $M (h, k)$ be the mid-point of the line segment PQ. Using the mid-point formula:

h = \frac{t_{1}^{2} + t_{2}^{2}}{2} and k = \frac{2 t_{1} + 2 t_{2}}{2} = t_{1} + t_{2}

From $k = t_{1} + t_{2}$ , we square both sides to get $k^{2} = (t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$ . Substitute $t_{1} t_{2} = - 4$ from Step 1:

k^{2} = t_{1}^{2} + t_{2}^{2} + 2 (- 4) ⟹ t_{1}^{2} + t_{2}^{2} = k^{2} + 8

Now substitute this expression for $t_{1}^{2} + t_{2}^{2}$ into the equation for $h$ :

h = \frac{k^{2} + 8}{2} ⟹ 2 h = k^{2} + 8

Replacing $(h, k)$ with $(x, y)$ to find the locus, we get the equation of conic C:

y^{2} = 2 x - 8 ⟹ y^{2} = 2 (x - 4)

💡 Teacher's Secret Hint

Remember to eliminate the parameters $t_{1}$ and $t_{2}$ to find the locus equation in terms of $x$ and $y$ .

Step 3: Identify Conic and Calculate Latus Rectum○ Expand

The equation $y^{2} = 2 (x - 4)$ is of the form $Y^{2} = 4 A X$ , where $Y = y$ and $X = x - 4$ . This represents a parabola. Comparing $y^{2} = 2 (x - 4)$ with the standard form $Y^{2} = 4 A X$ , we have $4 A = 2$ . The length of the latus rectum of a parabola $Y^{2} = 4 A X$ is $4 A$ . Therefore, the length of the latus rectum of conic C is:

Length of Latus Rectum = 4 A = 2

💡 Teacher's Secret Hint

Ensure you correctly identify the standard form of the conic and the corresponding formula for its latus rectum.

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Geometric Properties

🧠 Full Solution Path