Let the image of parabola x² = 4y, in the line x - y = 1 be (y + a)² = b(x - c), a, b, c ∈ N. Then a

Q: Why can't we just reflect the vertex and the focus to find the new parabola?

While reflecting the vertex and focus is a valid geometric approach, it is often prone to errors because you must also ensure the axis of symmetry is correctly transformed. The locus method (reflecting a general point) is more robust because it automatically accounts for the transformation of the entire curve, including its orientation and axis.

Q: Why do we use $(h, k)$ for the original point and $(X, Y)$ for the image?

This is a crucial habit in coordinate geometry! Using different variables helps you distinguish between the coordinates of the original curve and the coordinates of the transformed image. It prevents you from accidentally substituting the image coordinates back into the original equation prematurely.

Q: Is the reflection of a parabola always another parabola?

Yes, absolutely! Reflection is an isometry, meaning it preserves distances and angles. Since a parabola is defined by its geometric properties (like the set of points equidistant from a focus and a directrix), and reflection preserves these properties, the image of a parabola under reflection will always be another parabola of the same shape.

Question

Let the image of parabola $x^2 = 4y$, in the line $x - y = 1$ be $(y + a)^2 = b(x - c)$, $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to

Options:

(A) 8

(B) 6

(C) 12

(D) 4

Sigma Percentile · Accepted Answer

### The Art of Reflection: A Geometric Journey

Welcome, fellow traveler in the world of coordinate geometry! Today, we are going to tackle a problem that might look intimidating at first glance: reflecting a parabola across a line.

But fear not! We are not going to brute-force this. We are going to use the elegance of the locus method to unveil the hidden symmetry of the problem.

### Phase 1: The Locus Strategy

Imagine you are standing on the curve of the parabola $x^2 = 4y$. Every point $(h, k)$ on this curve obeys the sacred rule:

$$h^2 = 4k$$

Now, imagine a mirror placed along the line $x - y = 1$. When we reflect the entire parabola, we are essentially asking: where does every point $(h, k)$ land on the other side of the mirror?

Instead of trying to reflect the whole shape, we pick a general point $(h, k)$ and find its image $(X, Y)$. The path traced by this image point $(X, Y)$ will be our new parabola.

### Phase 2: The Mirror Math

To find the image $(X, Y)$ of a point $(h, k)$ across the line $x - y - 1 = 0$, we use the reflection formula. For a line $Ax + By + C = 0$, the image $(X, Y)$ is given by:

$$\frac{X - h}{A} = \frac{Y - k}{B} = -2 \frac{Ah + Bk + C}{A^2 + B^2}$$

Here, $A = 1$, $B = -1$, and $C = -1$. Substituting these values, we get:

$$\frac{X - h}{1} = \frac{Y - k}{-1} = -2 \frac{h - k - 1}{1^2 + (-1)^2}$$

The denominator simplifies beautifully: $1^2 + (-1)^2 = 2$. This cancels the $-2$ in the numerator, leaving us with:

$$\frac{X - h}{1} = \frac{Y - k}{-1} = -(h - k - 1) = k - h + 1$$

### Phase 3: The Inverse Mapping

Now, we solve for $X$ and $Y$. From $X - h = k - h + 1$, we get $X = k + 1$.

From $-(Y - k) = k - h + 1$, we get $Y - k = h - k - 1$, which simplifies to $Y = h - 1$. We have our mapping!

But we need the inverse to map back to the original curve: $k = X - 1$ and $h = Y + 1$. This is the "Aha!" moment.

We know that the original point $(h, k)$ must satisfy $h^2 = 4k$. By substituting our inverse mappings, we get:

$$(Y + 1)^2 = 4(X - 1)$$

### The Final Reveal

Replacing $(X, Y)$ with general coordinates $(x, y)$, we get the equation of the image parabola:

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

Comparing this with the given form $(y + a)^2 = b(x - c)$, we find $a = 1$, $b = 4$, and $c = 1$.

The sum is:

$$a + b + c = 1 + 4 + 1 = \mathbf{6}$$

And there you have it—a beautiful, logical journey from a simple parabola to its reflected twin. Keep practicing, and remember: every complex problem is just a series of simple, elegant steps waiting to be discovered!

Animated Solution for Mathematics - Conic Sections: Let the image of parabola $x^{2} = 4 y$ , in the line $x - y = 1$ be $(y + a)^{2} = b (x - c)$ , $a, b, c \in N$ . Then $a + b + c$ is equal to

Select Answer:

Visualized Solution

Visualizing the Setup

The Reflection Strategy

The Reflection Formula

Substituting Line Parameters

Simplifying the Expression

Finding Image Coordinates

Inverse Mapping

Applying the Original Constraint

The Image Parabola

Comparing with Given Form

Extracting Constants

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Art of Reflection

Phase 1

Phase 2

Phase 3

The Final Reveal

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the image of parabola x2=4y, in the line x−y=1 be (y+a)2=b(x−c), a,b,c∈N. Then a+b+c is equal to