Let O be the vertex of the parabola x² = 4y and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2:3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1,2), i

The parabola $x^2 = 4ay$ has vertex at origin, opens upward, with focus at $(0,a)$.

$x^2 = 4ay$ → parametric: $(2at,\ at^2)$ $y^2 = 4ax$ → parametric: $(at^2,\ 2at)$

For $x^2 = 4y$, we have $a=1$, so $Q = (2t, t^2)$. Always verify: $(2t)^2 = 4t^2 = 4(t^2)$ ✓

If P divides segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in ratio $m:n$ internally: In this problem: $O=(0,0)$, $Q=(2t,t^2)$, ratio $= 2:3$:
$h = \frac{2(2t)+3(0)}{5} = \frac{4t}{5}$,   $k = \frac{2t^2}{5}$

After getting $h = \frac{4t}{5}$ and $k = \frac{2t^2}{5}$:

Step 1: Express $t$ from the simpler equation: $t = \frac{5h}{4}$
Step 2: Substitute into the other: $k = \frac{2}{5}\cdot\frac{25h^2}{16} = \frac{5h^2}{8}$
Step 3: Write in standard form: $5h^2 = 8k$ → replace $h,k$ with $x,y$

This gives conic $C: 5x^2 = 8y$ — a parabola with axis along y-axis.

This is the most important formula for chord-bisection problems in JEE:

For conic $S = 0$, the chord bisected at $(x_1, y_1)$ satisfies $T = S_1$ where:
• $T$ = equation obtained by replacing in $S$: $x^2 \to xx_1$, $y^2 \to yy_1$, $x \to \frac{x+x_1}{2}$, $y \to \frac{y+y_1}{2}$
• $S_1$ = value of $S$ at $(x_1, y_1)$

For $S = 5x^2 – 8y$:
$T = 5(x)(x_1) – 8\cdot\frac{y+y_1}{2} = 5x_1 x – 4(y+y_1)$
$S_1 = 5x_1^2 – 8y_1$

At $(1,2)$: $T = 5x – 4(y+2) = 5x-4y-8$ and $S_1 = 5-16=-11$
$T=S_1 \Rightarrow 5x-4y-8=-11 \Rightarrow 5x-4y+3=0$

Let the chord have endpoints $(x_1+h_1, y_1+k_1)$ and $(x_1-h_1, y_1-k_1)$ symmetric about midpoint $(x_1, y_1)$. Both points lie on the conic. Subtracting their conic equations eliminates the constant and gives a linear relation — which is exactly $T = S_1$. This is a powerful algebraic shortcut that avoids solving for the actual endpoints.

Key insight: The locus equation (conic $C$) must be found first — $T=S_1$ is applied to conic $C$, not the original parabola!

❌ Mistake 1: Applying $T=S_1$ to the original parabola $x^2=4y$ instead of the locus conic $C: 5x^2=8y$.

❌ Mistake 2: Getting the section formula ratio reversed — remember O comes first (coordinates $(0,0)$), Q comes second.

❌ Mistake 3: Error in computing $T$ for $5x^2-8y=0$. The $y$ term has coefficient $-8$, so the midpoint substitution gives $-8\cdot\frac{y+y_1}{2} = -4(y+y_1)$.

❌ Mistake 4: Not simplifying correctly: $5x-4y-8=-11 \Rightarrow 5x-4y+3=0$ (add 11 to both sides).

Step 1: Parametrize $Q$ on original conic.
Step 2: Use section formula to get $(h,k)$ in terms of parameter.
Step 3: Eliminate parameter → locus equation = conic $C$.
Step 4: Apply $T=S_1$ on conic $C$ with given midpoint.
Step 5: Simplify to get chord equation.

Total time in exam: ~3 minutes

This problem combines two classic JEE topics:
• Locus problems (section formula + parametric elimination)
• Chord with given midpoint ($T=S_1$ method)

Both are guaranteed topics in JEE Main every year. The $T=S_1$ formula for parabola, ellipse, and hyperbola is a must-memorise result. Locus problems involving section formula appear at least once per session. Together, this question tests 2 high-value concepts in one problem.

Parabola $x^2=4ay$: param $(2at, at^2)$ Section formula: $\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$ Chord midpoint: $T = S_1$ For $5x^2-8y=0$: $T = 5xx_1 – 4(y+y_1)$ For $y^2=4ax$: chord midpoint $T = yy_1-2a(x+x_1)$ For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: $T=\frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1$