JEE Mathematics 2026 Question with Solution

Given: The parabola is $x^2=4y$ with vertex O at $(0,0)$. A point Q lies on the parabola, and P divides OQ internally in the ratio $2:3$. Find: The equation of the chord of the conic C bisected at $(1,2)$.

A general point on $x^2=4y$ can be written in parametric form as $Q=(2t,t^2)$.

Since P divides OQ in the ratio $2:3$,

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

and

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

So,

$$t = \frac{5h}{4}$$

Substituting into the expression for $k$,

$$k = \frac{2}{5}\left(\frac{5h}{4}\right)^2 = \frac{2}{5}\cdot \frac{25h^2}{16} = \frac{5h^2}{8}$$

Hence the locus of P is

$$8k=5h^2$$

or, replacing $h,k$ by $x,y$,

$$5x^2-8y=0$$

Thus the conic is $C: 5x^2-8y=0$.

For the chord of a conic $S=0$ bisected at $(x_1,y_1)$, we use the formula $T=S_1$. Here,

$$S \equiv 5x^2-8y$$

and $(x_1,y_1)=(1,2)$.

Forming $T$ by replacing $x^2$ with $xx_1$ and $y$ with $\frac{y+y_1}{2}$,

$$T \equiv 5xx_1 - 8\left(\frac{y+y_1}{2}\right)$$ $$T = 5x(1) - 8\left(\frac{y+2}{2}\right) = 5x-4y-8$$

Now,

$$S_1 = 5(1)^2 - 8(2) = 5-16 = -11$$

Using $T=S_1$,

$$5x-4y-8=-11$$ $$5x-4y+3=0$$

Therefore, the equation of the required chord is $5x-4y+3=0$. The correct option is A.

Given: The required chord belongs to the locus of the point dividing OQ in the ratio $2:3$. Find: The chord of that conic bisected at $(1,2)$.

First find the locus quickly. If $Q=(2t,t^2)$ on $x^2=4y$, then the section formula gives

$$P=\left(\frac{4t}{5},\frac{2t^2}{5}\right)$$

Eliminating $t$ gives

$$5x^2-8y=0$$

Now use the shortcut for a chord of a conic bisected at $(x_1,y_1)$:

$$T=S_1$$

For $S\equiv 5x^2-8y$ at $(1,2)$,

$$T=5x(1)-8\left(\frac{y+2}{2}\right)=5x-4y-8$$

and

$$S_1=5(1)^2-8(2)=-11$$

So,

$$5x-4y-8=-11 \implies 5x-4y+3=0$$

This works because $T=S_1$ is the standard result for the chord of any second-degree conic bisected at a given point. Hence the correct option is A.