JEE Main 2025 (January)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let A and B be the two points of intersection of the line $y + 5 = 0$ and the mirror image of the parabola $y^{2} = 4 x$ with respect to the line $x + y + 4 = 0$ If d denotes the distance between A and B, and a denotes the area of $△ S A B,$ where S is the focus of the parabola $y^{2} = 4 x$ then the value of $(a + d)$ is

Q: Wait, why is the height of the triangle $|0 - (-5)| = 5$?

The focus $S$ of the original parabola $y^2=4x$ is at $(1, 0)$. The line of intersection is $y+5=0$, which is a horizontal line $y=-5$. The perpendicular distance from any point $(x_0, y_0)$ to a horizontal line $y=k$ is simply $|y_0 - k|$. Here, the y-coordinate of the focus is $0$, and the line is $y=-5$, so the height is $|0 - (-5)| = 5$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the symmetry. Instead of reflecting the entire parabola, consider how the mirror line

x + y + 4 = 0

acts as a geometric axis of symmetry. The key is to understand that the reflection of any point

(x, y)

across this line will map to a new point

(X, Y)

on the reflected parabola. Can you see how the parabola's orientation changes after the reflection?

Hint 2

Hint 2 (The Arsenal): You need the reflection formula for a point

(x, y)

across a line

a x + b y + c = 0

. The formula is

\frac{X - x}{a} = \frac{Y - y}{b} = - 2 \frac{a x + b y + c}{a ^{2} + b ^{2}}

. This is your primary tool to transform the original equation

y^{2} = 4 x

into the equation of the reflected parabola.

Hint 3

Hint 3 (The Execution): Once you have the equation of the reflected parabola, substitute

y = - 5

to find the intersection points

A

and

B

. The distance

d

is simply the absolute difference between the x-coordinates of

A

and

B

. For the area

a

△ S A B

, use the formula

\frac{1}{2} \times base \times height

, where the base is

d

and the height is the perpendicular distance from the focus

S (1, 0)

to the line

y = - 5

Visualized Solution

Analyze the Original Parabola $y^{2} = 4 x$

Original Parabola: $y^{2} = 4 x$
Standard Form: $y^{2} = 4 a x \Rightarrow 4 a = 4 \Rightarrow a = 1$
Focus $S = (a, 0) = (1, 0)$

The Mirror Line $x + y + 4 = 0$

Mirror Line: $x + y + 4 = 0$
We need to reflect the parabola $y^{2} = 4 x$ across this line.

Mirror Transformation Formula

Transformation Formula for image $(X, Y)$ of a point $(x, y)$ :
$\frac{X - x}{a} = \frac{Y - y}{b} = - 2 \frac{a x + b y + c}{a ^{2} + b ^{2}}$

Deriving $X$ and $Y$ in terms of Image

Here $a = 1, b = 1, c = 4$ .
$\frac{X - x}{1} = \frac{Y - y}{1} = - 2 \frac{x + y + 4}{1 ^{2} + 1 ^{2}}$
Simplifying gives $X = - y - 4$ and $Y = - x - 4$

Equation of the Mirrored Parabola

Substitute into $Y^{2} = 4 X$ :
$(- x - 4)^{2} = 4 (- y - 4)$
$(x + 4)^{2} = - 4 (y + 4)$

Intersection with Line $y + 5 = 0$

Intersection with line $y + 5 = 0 \Rightarrow y = - 5$

Solving for $x$ -coordinates

Substitute $y = - 5$ into $(x + 4)^{2} = - 4 (y + 4)$ :
$(x + 4)^{2} = - 4 (- 5 + 4) = - 4 (- 1) = 4$

Identifying Points $A$ and $B$

$(x + 4)^{2} = 4 \Rightarrow x + 4 = \pm 2$
Case 1: $x + 4 = 2 \Rightarrow x = - 2$
Case 2: $x + 4 = - 2 \Rightarrow x = - 6$
Points of intersection: $A (- 2, - 5)$ and $B (- 6, - 5)$

Calculating Distance $d$

Distance $d = A B = ∣ - 2 - (- 6) ∣ = 4$

Setup for Area of $△ S A B$

Triangle vertices: $S (1, 0), A (- 2, - 5), B (- 6, - 5)$
Base $d = 4$
Height $h = ∣0 - (- 5) ∣ = 5$

Calculating Area $a$

Area $a = \frac{1}{2} \times 4 \times 5 = 10$

Final Sum $a + d$

Value of $a + d = 10 + 4 = 14$
Final Answer: 14

00:00 / 00:00

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

Solution Diagram

Analyzing the Geometry of the Parabola

We begin with the classic parabola defined by the equation

y^{2} = 4 x

. In the standard form

y^{2} = 4 a x

, we identify

a = 1

This immediately reveals that the focus

S

is located at

(1, 0)

. This point serves as the anchor for the entire curve.

The Mirror Transformation

We reflect the parabola across the line

x + y + 4 = 0

. To find the new equation, we utilize the reflection formula for a point

(x, y)

across a line

a x + b y + c = 0

\frac{X - x}{a} = \frac{Y - y}{b} = - 2 \frac{a x + b y + c}{a ^{2} + b ^{2}}

Substituting

a = 1

b = 1

, and

c = 4

, we derive the transformation equations:

X = - y - 4

Y = - x - 4

By substituting these into the original equation

Y^{2} = 4 X

, we obtain the equation of the reflected parabola:

(- x - 4)^{2} = 4 (- y - 4)

Simplifying this expression, we arrive at the equation of the reflected curve:

(x + 4)^{2} = - 4 (y + 4)

The Intersection

The problem requires finding the intersection of this new parabola with the line

y + 5 = 0

, which is the horizontal line

y = - 5

. Substituting

y = - 5

into our reflected equation:

(x + 4)^{2} = - 4 (- 5 + 4)

(x + 4)^{2} = - 4 (- 1) = 4

This yields

x + 4 = \pm 2

. Solving for

x

, we find

x = - 2

and

x = - 6

Thus, the intersection points are

A (- 2, - 5)

and

B (- 6, - 5)

Final Calculation

The distance

d

between points

A

and

B

is the difference in their

x

-coordinates:

d = ∣ - 2 - (- 6) ∣ = 4

Next, we calculate the area

a

△ S A B

. The base of the triangle is

d = 4

, and the height is the vertical distance from the focus

S (1, 0)

to the line

y = - 5

Height = ∣0 - (- 5) ∣ = 5

The area

a

is given by:

a = \frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times 5 = 10

Finally, the value of

a + d

is:

10 + 4 = 14

Common Traps & FAQs

Why do we need to reflect the entire parabola? Can't we just reflect the points?

Reflecting the entire curve is the most robust method. If you only reflect points, you would not know which points on the original parabola correspond to the intersection points on the reflected one. By finding the equation of the reflected parabola, you create a general tool that works for any line intersecting it.

Is there a faster way to find the reflected parabola without using the full point-reflection formula?

While you could use properties of the focus and directrix, the point-reflection formula is the most reliable and standard approach in JEE Advanced. It minimizes the chance of conceptual errors and is universally applicable to any line of reflection.

Wait, why is the height of the triangle

∣0 - (- 5) ∣ = 5

The focus

S

of the original parabola

y^{2} = 4 x

is at

(1, 0)

. The line of intersection is

y + 5 = 0

, which is a horizontal line

y = - 5

. The perpendicular distance from any point

(x_{0}, y_{0})

to a horizontal line

y = k

is simply

∣ y_{0} - k ∣

. Here, the y-coordinate of the focus is

0

, and the line is

y = - 5

, so the height is

∣0 - (- 5) ∣ = 5

Enter Numerical Value:

Visualized Solution

Analyze the Original Parabola $y^{2} = 4 x$

The Mirror Line $x + y + 4 = 0$

Mirror Transformation Formula

Deriving $X$ and $Y$ in terms of Image

Equation of the Mirrored Parabola

Intersection with Line $y + 5 = 0$

Solving for $x$ -coordinates

Identifying Points $A$ and $B$

Calculating Distance $d$

Setup for Area of $△ S A B$

Calculating Area $a$

Final Sum $a + d$

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

Analyzing the Geometry of the Parabola

The Mirror Transformation

The Intersection

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Analyze the Original Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4x

The Mirror Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y+4=0

Mirror Transformation Formula

Deriving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y in terms of Image

Equation of the Mirrored Parabola

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

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

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

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

Setup for Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △SAB

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

Final Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+d

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

Analyzing the Geometry of the Parabola

The Mirror Transformation

The Intersection

Final Calculation

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Final Sum $a + d$

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