JEE Mathematics 2025 Question with Solution

Given: End $A$ lies on $x-y+2=0$ and end $B$ lies on $y+2=0$. The rod has fixed length $8$ units. Point $P(x,y)$ divides $AB$ internally in the ratio $2:1$.

Find: $\alpha-\beta-\gamma$ from the locus of $P$.

Take

since $A$ lies on $x-y+2=0$ and $B$ lies on $y=-2$.

Using the length of the rod,

so

Since $P$ divides $AB$ internally in the ratio $2:1$,

Hence,

From $y=\frac{x_1-2}{3}$, we get

And from $x=\frac{2x_2+x_1}{3}$,

so

Now,

and

Substitute in the length condition:

Multiply by $4$:

Divide by $4$ form equivalently to match the given locus after expansion.

Expanding,

Factor out $9$ from the required part:

Comparing with

we get

Therefore,

So, the correct option is B.

Given: $A$ lies on $x-y+2=0$, $B$ lies on $y=-2$, and $AB=8$.

Find: The value of $\alpha-\beta-\gamma$.

Write the moving points as

Then

If $P(x,y)$ divides $AB$ in the ratio $2:1$, then

So

Substitute these into

This gives

Now simplify:

Expanding and collecting terms,

Hence,

Therefore,

and

Hence the correct option is B.