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Parabola: Parabola Reflected Line Image Find

JEE Maths question with a full step-by-step solution.

Question
The parabola y=x2y = x^{2} is reflected in the line x+y=1x + y = 1, and the image is x+y2=kyx + y^{2} = ky. Find kk.
Solution
Answer: 2
Question attachment Step 1: Let A(t, t2)A\left(t,\,t^{2}\right) be a general point of the given parabola and let B(a,b)B(a,b) be its mirror image in the line x+y=1x+y=1. Step 2: The mid point of ABAB lies on the mirror line:
a+t2+b+t22=1 ⟹ a+t+b+t2=2(1)\frac{a+t}{2} + \frac{b+t^{2}}{2} = 1 \ \Longrightarrow\ a + t + b + t^{2} = 2 \tag{1}
 Step 3: ABAB is perpendicular to the mirror line. The line x+y=1x+y=1 has slope −1-1, so ABAB has slope 11:
b−t2a−t=1 ⟹ b−t2=a−t(2)\frac{b-t^{2}}{a-t} = 1 \ \Longrightarrow\ b - t^{2} = a - t \tag{2}
 If AA itself lies on the mirror line then B=AB=A and (2)(2) holds all the same. Step 4: Writing (2)(2) as b−a=t2−tb - a = t^{2} - t and adding it to (1)(1),
(a+b)+(b−a)=(2−t−t2)+(t2−t)=2−2t ⟹ 2b=2−2t ⟹ t=1−b(a+b) + (b-a) = \left(2 - t - t^{2}\right) + \left(t^{2}-t\right) = 2 - 2t \ \Longrightarrow\ 2b = 2-2t \ \Longrightarrow\ t = 1-b
 Step 5: From (1)(1), a=2−t−t2−ba = 2 - t - t^{2} - b, and using b=1−tb = 1-t,
a=2−t−t2−(1−t)=1−t2=1−(1−b)2=2b−b2a = 2 - t - t^{2} - (1-t) = 1 - t^{2} = 1 - (1-b)^{2} = 2b - b^{2}
 Step 6: Replacing (a,b)(a,b) by (x,y)(x,y),
x=2y−y2 ⟹ x+y2=2yx = 2y - y^{2} \ \Longrightarrow\ x + y^{2} = 2y
 Comparing with x+y2=kyx + y^{2} = ky gives k=2k = 2. Step 7: Check with one point. The vertex (0,0)(0,0) of y=x2y=x^{2} reflects in x+y=1x+y=1 to (1,1)(1,1), and (1,1)(1,1) satisfies x+y2=2yx+y^{2}=2y, since 1+1=21+1 = 2. Answer: 22.
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