JEE Mathematics 2026 Question with Solution

Given: The parabola is $y^2 = 16x$, so comparing with $y^2 = 4ax$ gives $a = 4$. One end of a focal chord is $A(16,16)$.

Find: The minimum value of $\alpha + \beta$, where $P(\alpha,\beta)$ divides the focal chord internally in the ratio $5:2$.

For a point on the parabola $y^2 = 4ax$ in parametric form, the coordinates are

Since $A(16,16) = (4t_1^2, 8t_1)$, from $8t_1 = 16$ we get

Because the chord is a focal chord, the parameters of its endpoints satisfy

Hence

So the other end $B$ is

Now $P(\alpha,\beta)$ divides $AB$ internally in the ratio $5:2$. Two internal orders are possible.

Case 1: Ratio $5:2$ from $A$ to $B$. Using the section formula,

Therefore,

Case 2: Ratio $5:2$ from $B$ to $A$. Then

So,

The minimum of the two values is $7$. Therefore, the correct option is B.

Given: One endpoint of a focal chord of $y^2 = 16x$ is $(16,16)$.

Find: The least value of $\alpha + \beta$ for the internal division point.

For $y^2 = 4ax$, the focus is $(a,0)$ and any point with parameter $t$ is

Here $a = 4$, so the point corresponding to parameter $t$ is

Matching with $(16,16)$ gives

The endpoints of a focal chord satisfy

Thus

Hence the second endpoint is

Now apply the internal section formula carefully, because the statement $5:2$ can be interpreted in either order along the segment.

If $AP:PB = 5:2$, then

Substituting $A(16,16)$ and $B(1,-4)$,

So

If $BP:PA = 5:2$, then

Therefore

Thus

Among the possible internal division points, the minimum value of $\alpha + \beta$ is $7$. Hence the correct option is B.