Question

Let mirror image of parabola $x^2 = 4y$ in the line $x-y=1$ be $(y+a)^2 = b(x-c)$. Then the value of $(a+b+c)$ is

Hint:

For mirror image problems, parametric points simplify reflection calculations greatly.

Answer 1

Correct Answer: 6

Step 1: Choose a general point on the parabola

The given parabola is

x² = 4y

A general point on it can be written in parametric form as:

P(2t, t²)

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Step 2: Use the reflection formula about the line x − y = 1

The mirror image of a point (x₁, y₁) in the line

ax + by + c = 0

is given by:

( x₁ − 2a(ax₁ + by₁ + c)/(a² + b²),   y₁ − 2b(ax₁ + by₁ + c)/(a² + b²) )

Here, x − y − 1 = 0 ⇒ a = 1, b = −1, c = −1

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Step 3: Find the reflected point

For P(2t, t²):

ax + by + c = 2t − t² − 1

a² + b² = 2

Thus, the reflected point Q is:

Q = ( 2t − (2(2t − t² − 1))/2,   t² + (2(2t − t² − 1))/2 )

Simplifying,

Q = (t² + 1, 2t − 1)

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Step 4: Eliminate the parameter

From y = 2t − 1,

t = (y + 1)/2

Substitute into x = t² + 1:

x = (y + 1)²/4 + 1

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Step 5: Obtain the equation of the reflected curve

Rewriting,

(y + 1)² = 4(x − 1)

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Step 6: Compare with the standard form

Comparing with

(y + a)² = b(x − c)

we get:

a = 1,   b = 4,   c = 1

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Step 7: Required sum

a + b + c = 1 + 4 + 1

= 6

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Final Answer:

The value of (a + b + c) is
6