Class JEE Mathematics ALL Q #1179
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
If O is the vertex of the parabola $x^{2}=4y,$ Q is a point on the parabola. If C is the locus of the point which divides OQ in the ratio $2:3,$ then the equation of the chord of C which is bisected at the point (1,2) is: (Duplicate Question)
(A) $5x+4y+3=0$
(B) $5x-4y-3=0$
(C) $5x-4y+3=0$
(D) $5x+4y-3=0$
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AI Tutor Explanation

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Step-by-Step Solution

Let Q be $(2t, t^2)$ on the parabola $x^2 = 4y$.

Let P(h, k) divide OQ in the ratio 2:3. Then,

$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5}$ and $k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5}$

So, $t = \frac{5h}{4}$ and $k = \frac{2}{5} (\frac{5h}{4})^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$

Therefore, the locus of C(h, k) is $5x^2 = 8y$.

The equation of the chord of $5x^2 = 8y$ bisected at (1, 2) is given by $T = S_1$, where $T = 5x(1) - 4(y+2)$ and $S_1 = 5(1)^2 - 8(2)$.

So, $5x - 4y - 8 = 5 - 16 = -11$

$5x - 4y - 8 + 11 = 0$

$5x - 4y + 3 = 0$

Correct Answer: $5x-4y+3=0$

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AI Suggestion: Option C

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply their knowledge of parabolas, section formula, and finding the equation of a chord bisected at a given point to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps to arrive at the solution, including using the section formula, finding the locus, and applying the formula for the equation of a chord bisected at a point.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires application of multiple concepts related to coordinate geometry and parabolas, going beyond simple recall or understanding of definitions.

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