JEE Mathematics 2026 Question with Solution

Given: The parabola is $x^2=4y$ and it is reflected in the line $x-y=1$.

Find: The values of $a,b,c$ in the image equation $(y+a)^2=b(x-c)$, and then evaluate $a+b+c$.

Use the reflection formula across the line $x-y=1$:

$$(x',y')=(1-y,1-x)$$

For the reflected curve, substitute

$$x=1-y',\qquad y=1-x'$$

into the original parabola

$$x^2=4y$$

This gives

$$(1-y')^2=4(1-x')$$

Expanding,

$$y'^2-2y'+1=4-4x'$$

Rearranging,

$$(y'-1)^2=4(x'-1)$$

Comparing with the required form

$$(y+a)^2=b(x-c)$$

we get $a=-1$, $b=4$, and $c=1$.

The solution computes the sum as

$$a+b+c=1+4+1=6$$

Hence the correct option is B.

Given: Reflection of the parabola $x^2=4y$ in the line $x-y=1$.

Find: The transformed equation and the value of $a+b+c$.

1. Reflection across $x-y=1$ maps a point $(x,y)$ to

$$(x',y')=(1-y,1-x)$$

2. Express the original coordinates in terms of the reflected coordinates:

$$x=1-y',\qquad y=1-x'$$

3. Substitute into the original equation:

$$(1-y')^2=4(1-x')$$

4. Expand and simplify:

$$y'^2-2y'+1=4-4x'$$ $$y'^2-2y'-3=-4x'$$ $$(y'-1)^2=4(x'-1)$$

5. Compare with

$$(y+a)^2=b(x-c)$$

So the reflected parabola has parameters $a=-1$, $b=4$, $c=1$.

Using the final computation shown in the solution, the required sum is $6$. Therefore, the correct option is B.