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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Parabola

Let the image of parabola x^ 2 =4 y , in the line x-y=1 be (y+a)^ 2...

Let the image of parabola x2=4yx^{2}=4 y, in the line x−y=1x-y=1 be (y+a)2=b(x−c)(y+a)^{2}=b(x-c), a,b,c∈ Na, b, c \in \mathrm{~N}. Then a+b+ca+b+c is equal to

Options

  1. A8
  2. B6
  3. C12
  4. D4

Correct answer

B. 6

Step-by-step solution

Reflection of point (h,k)(h, k) in line x−y−1=0x - y - 1 = 0:
x−h1=y−k−1=−2(h−k−1)2=k−h+1\frac{x-h}{1} = \frac{y-k}{-1} = \frac{-2(h-k-1)}{2} = k - h + 1.
Image: (k+1, h−1)(k+1,\ h-1).
Let (X,Y)=(k+1, h−1)(X, Y) = (k+1,\ h-1), so h=Y+1, k=X−1h = Y+1,\ k = X-1.
From h2=4kh^2 = 4k: (Y+1)2=4(X−1)(Y+1)^2 = 4(X-1).
Comparing with (y+a)2=b(x−c)(y+a)^2 = b(x-c): a=1, b=4, c=1a = 1,\ b = 4,\ c = 1.
a+b+c=1+4+1=6a + b + c = 1 + 4 + 1 = 6.
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