JEE Mathematics 2026 Question with Solution

Given: The parabola is $y^2=8x$, the focus is $A$, the line intersects the parabola at two distinct points $B(x_1,y_1)$ and $C(x_2,y_2)$, and the centroid of triangle $ABC$ is $\left(\frac{7}{3},\frac{4}{3}\right)$.

Find: The value of $(BC)^2$.

For $y^2=4ax$, the focus is $(a,0)$. Here,

$$4a=8$$

so,

$$a=2$$

Therefore,

$$A=(2,0)$$

Using the centroid formula for triangle $ABC$,

$$\left(\frac{x_1+x_2+2}{3},\frac{y_1+y_2}{3}\right)=\left(\frac{7}{3},\frac{4}{3}\right)$$

Thus,

$$x_1+x_2=5$$

and

$$y_1+y_2=4$$

Now,

$$(BC)^2=(x_1-x_2)^2+(y_1-y_2)^2$$

Using identities,

$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$$

and

$$(y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2$$

Since the points lie on the parabola $y^2=8x$,

$$y_1^2+y_2^2=8(x_1+x_2)=40$$

Also,

$$(y_1+y_2)^2=y_1^2+y_2^2+2y_1y_2$$

So,

$$16=40+2y_1y_2$$

which gives

$$y_1y_2=-12$$

Now,

$$x_1x_2=\frac{y_1^2y_2^2}{64}=\frac{144}{64}=\frac{9}{4}$$

Therefore,

$$(x_1-x_2)^2=25-4\cdot \frac{9}{4}=25-9=16$$

and

$$(y_1-y_2)^2=16-4(-12)=16+48=64$$

Hence,

$$(BC)^2=16+64=80$$

the solution shows an arithmetic discrepancy in the final computation, but the correct evaluation from the extracted steps is $80$. Therefore, the correct option is D.

Given: $x_1+x_2=5$ and $y_1+y_2=4$ from the centroid condition.

Find: Recompute $(BC)^2$ carefully.

From

$$y_1y_2=-12$$

we get

$$(y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2=16-4(-12)=16+48=64$$

Also,

$$x_1x_2=\frac{9}{4}$$

so

$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=25-9=16$$

Thus,

$$(BC)^2=(x_1-x_2)^2+(y_1-y_2)^2=16+64=80$$

Hence the listed conclusion $32$ in the source working is inconsistent with its own intermediate results. The defensible answer is D.