Let the curve C be the mirror image of the parabola y² = 4x with respect to the line x + y + 4 = 0.

Q: Wait, why do we reject the idea of finding the equation of the reflected curve directly?

You *could* find the equation, but it is like taking the long way around a mountain. It involves eliminating the parameter $t$ and rearranging complex algebraic terms, which is a breeding ground for sign errors. Using parametric mapping is elegant, keeps the algebra clean, and is the standard, efficient approach for JEE-level coordinate geometry.

Q: How do I know the reflection formula is reliable?

The reflection formula is derived from two fundamental geometric truths: the line segment connecting the point and its image is perpendicular to the mirror line, and the midpoint of that segment lies on the mirror line. It is a robust, time-tested tool. Trust the formula, but always double-check your coefficients $a$, $b$, and $c$!

Q: What if the intersection line was not horizontal? Would the distance calculation change?

Absolutely! If the intersection line were slanted, you would need to use the full distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Here, because the intersection line $y = -5$ is horizontal, the $y$-coordinates are identical, simplifying the distance to just the absolute difference of the $x$-coordinates. Always look for these geometric shortcuts to save precious time.

Question

Let the curve $C$ be the mirror image of the parabola $y^2 = 4x$ with respect to the line $x + y + 4 = 0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y = -5$, then the distance between $A$ and $B$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mirror, but instead of seeing yourself, you see the elegant geometry of a parabola. This problem is not just about finding a distance; it is about mastering the art of transformation.

We are given a parabola $y^2 = 4x$ and a mirror line $x + y + 4 = 0$. Our mission is to find the distance between the intersection points of the reflected curve $C$ and the horizontal line $y = -5$.

### The Parametric Strategy

The most common mistake students make is trying to derive the full Cartesian equation of the reflected curve. That path is fraught with algebraic peril.

Instead, let us embrace the power of parameters. By representing any point on the parabola as $P(t^2, 2t)$, we reduce the entire curve to a single variable $t$.

This is the secret weapon of coordinate geometry—it allows us to track the movement of the entire curve by simply tracking the movement of a single, general point.

### The Reflection Formula

Now, we invoke the reflection formula. For a point $(x_1, y_1)$ reflected across the line $ax + by + c = 0$, the image $(h, k)$ is determined by the ratio:

$$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Substituting our point $P(t^2, 2t)$ and the line $x + y + 4 = 0$ (where $a=1, b=1, c=4$), we get:

$$\frac{h - t^2}{1} = \frac{k - 2t}{1} = \frac{-2(t^2 + 2t + 4)}{1^2 + 1^2}$$

Notice the elegance here: the denominator $1^2 + 1^2$ equals $2$, which perfectly cancels the $-2$ in the numerator. We are left with:

$$\frac{h - t^2}{1} = \frac{k - 2t}{1} = -(t^2 + 2t + 4)$$

### Solving for the Reflected Coordinates

With the formula simplified, we can isolate $h$ and $k$. For the $x$-coordinate:

$$h - t^2 = -(t^2 + 2t + 4) \Rightarrow h = -2t - 4$$

Similarly, for the $y$-coordinate:

$$k - 2t = -(t^2 + 2t + 4) \Rightarrow k = -t^2 - 4$$

We have now successfully mapped every point on the original parabola to its reflected counterpart on curve $C$ using only the parameter $t$.

### The Intersection and Final Distance

The problem asks for the intersection with $y = -5$. Since our reflected $y$-coordinate is $k = -t^2 - 4$, we simply set:

$$-t^2 - 4 = -5$$

This yields $t^2 = 1$, meaning $t = 1$ or $t = -1$. These two values of $t$ give us the two points of intersection, $A$ and $B$.

For $t = 1$, $h = -2(1) - 4 = -6$, giving $A(-6, -5)$. For $t = -1$, $h = -2(-1) - 4 = -2$, giving $B(-2, -5)$.

Since both points lie on the line $y = -5$, the distance is simply the absolute difference between their $x$-coordinates:

$$|-2 - (-6)| = 4$$

We have navigated the reflection, solved the intersection, and arrived at the **final answer of 4**. Keep this parametric mindset, and no coordinate geometry problem will ever intimidate you again.

Animated Solution for Mathematics - Conic Sections: Let the curve $C$ be the mirror image of the parabola $y^{2} = 4 x$ with respect to the line $x + y + 4 = 0$ . If $A$ and $B$ are the points of intersection of $C$ with the line $y = - 5$ , then the distance between $A$ and $B$ is

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$P a r am e t r i c P o in t o n y^{2} = 4 x$

$T h e R e f l ec t i o n F or m u l a$

$A ppl y in g t h e F or m u l a$

$S im pl i f y in g t h e R e f l ec t i o n E q u a t i o n$

$S o l v in g f or x - coor d ina t e h$

$S o l v in g f or y - coor d ina t e k$

$I n t er sec t i o n w i t h y = - 5$

$F in d in g t h e P a r am e t er t$

$C oor d ina t es o f A an d B$

$C a l c u l a t in g t h e D i s t an ce A B$

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

Analyzing the Setup

The Parametric Strategy

The Reflection Formula

Solving for the Reflected Coordinates

The Intersection and Final Distance

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parametric Point on y2=4x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Reflection Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Reflection Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for x−coordinate h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for y−coordinate k

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Intersection with y=−5

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Parameter t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinates of A and B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Distance AB