JEE Main 2022 (29 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the focal chord of the parabola $P : y^{2} = 4 x$ along the line $L : y = m x + c, m > 0$ meet the parabola at the points $M$ and $N$ . Let the line $L$ be a tangent to the hyperbola $H : x^{2} - y^{2} = 4$ . If $O$ is the vertex of $P$ and $F$ is the focus of $H$ on the positive x-axis, then the area of the quadrilateral $O M F N$ is :

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the geometry. A 'focal chord' of a parabola is not just any line; it is a line that is strictly constrained to pass through the focus of the parabola. Since the parabola is

y^{2} = 4 x

, identify its focus

S

first. This gives you a fixed point

(1, 0)

that your line

L

must satisfy, effectively reducing the line equation

y = m x + c

to a single variable

m

Hint 2

Hint 2: The Arsenal. You are dealing with a tangency condition. When a line

y = m x + c

is tangent to a hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, there is a specific algebraic relationship between

m, c, a,

and

b

. Recall the condition

c^{2} = a^{2} m^{2} - b^{2}

. Apply this to your hyperbola

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

to find the value of

m

Hint 3

Hint 3: The Execution. Once you have the line equation, find the intersection points

M

and

N

by substituting the line into the parabola's equation. You will get a quadratic in

y

. Do not solve for

y

individually! Use the sum and product of roots to find the vertical distance

∣ y_{1} - y_{2} ∣

using the identity

∣ y_{1} - y_{2} ∣ = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2}

. Finally, calculate the area of the quadrilateral

O M F N

using the base-height formula, noting that

O

and

F

lie on the x-axis.

Visualized Solution

Identify the Parabola $P$

Parabola $P : y^{2} = 4 x$
Vertex $O (0, 0)$
Comparing with $y^{2} = 4 a x \Rightarrow a = 1$
Focus of $P$ is $S (1, 0)$

Equation of Focal Chord $L$

Line $L : y = m x + c$ passes through $S (1, 0)$
$0 = m (1) + c \Rightarrow c = - m$
Equation of $L : y = m (x - 1)$

Hyperbola $H$ Parameters

Hyperbola $H : x^{2} - y^{2} = 4$
Standard form: $\frac{x ^{2}}{4} - \frac{y ^{2}}{4} = 1$
$a^{2} = 4, b^{2} = 4$

Tangency Condition

Line $L$ is tangent to $H$
Condition: $c^{2} = a^{2} m^{2} - b^{2}$
Substitute $c = - m, a^{2} = 4, b^{2} = 4$ :
$(- m)^{2} = 4 m^{2} - 4$

Solve for Slope $m$

$m^{2} = 4 m^{2} - 4 \Rightarrow 3 m^{2} = 4$
$m^{2} = \frac{4}{3}$
Given $m > 0 \Rightarrow m = \frac{2}{3}$
Line $L : y = \frac{2}{3} (x - 1)$

Focus $F$ of Hyperbola

Eccentricity $e = 1 + \frac{b ^{2}}{a ^{2}} = 1 + \frac{4}{4} = 2$
Focus $F$ on positive x-axis: $(a e, 0)$
$F = (22, 0)$

Intersection Points $M$ and $N$

Intersection of $L : x = \frac{3}{2} y + 1$ and $P : y^{2} = 4 x$
$y^{2} = 4 (\frac{3}{2} y + 1)$
$y^{2} = 23 y + 4 \Rightarrow y^{2} - 23 y - 4 = 0$

Quadratic in $y$

For $y^{2} - 23 y - 4 = 0$ :
Sum of roots: $y_{1} + y_{2} = 23$
Product of roots: $y_{1} y_{2} = - 4$

Difference of y-coordinates

$∣ y_{1} - y_{2} ∣ = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2}$
$∣ y_{1} - y_{2} ∣ = (23)^{2} - 4 (- 4)$
$∣ y_{1} - y_{2} ∣ = 12 + 16 = 27$

Area of Quadrilateral $O M F N$

Vertices: $O (0, 0), M (x_{1}, y_{1}), F (22, 0), N (x_{2}, y_{2})$
Since $O$ and $F$ lie on x-axis:
Area $= \frac{1}{2} \times O F \times ∣ y_{1} - y_{2} ∣$
Area $= \frac{1}{2} (x_{F}) ∣ y_{1} - y_{2} ∣$

Final Calculation

Area $= \frac{1}{2} (22) (27)$
Area $= 2 \times 27$
Area $= 214$ square units

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

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are orchestrating a meeting between three distinct geometric entities: a parabola, a line, and a hyperbola.

Our first player is the parabola

P

, defined by the equation

y^{2} = 4 x

. By comparing this to the standard form

y^{2} = 4 a x

, we identify that

a = 1

. This tells us that the vertex

O

is at

(0, 0)

and the focus

S

is at

(1, 0)

The problem introduces a line

L

that acts as a focal chord passing through

S (1, 0)

. When we write the equation of a line as

y = m x + c

, the condition that it passes through

(1, 0)

forces

0 = m (1) + c

, which gives

c = - m

Thus, our line is defined by the parameter

m

as:

y = m (x - 1)

The Dance of Tangency

Next, we encounter the hyperbola

H : x^{2} - y^{2} = 4

. Dividing by

4

, we normalize it to:

\frac{x ^{2}}{4} - \frac{y ^{2}}{4} = 1

This is a rectangular hyperbola where

a^{2} = 4

and

b^{2} = 4

The problem states that line

L

is a tangent to this hyperbola. For a hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, the condition for the line

y = m x + c

to be a tangent is:

c^{2} = a^{2} m^{2} - b^{2}

Substituting our known values

c = - m

a^{2} = 4

, and

b^{2} = 4

, we get:

(- m)^{2} = 4 m^{2} - 4

3 m^{2} = 4 \Rightarrow m^{2} = \frac{4}{3}

Given

m > 0

, we find the slope to be

m = \frac{2}{3}

. Consequently, the line equation is

y = \frac{2}{3} (x - 1)

Intersection and Area Calculation

To find the intersection points

M

and

N

of the line and the parabola, we substitute

x = \frac{3}{2} y + 1

into

y^{2} = 4 x

y^{2} = 4 (\frac{3}{2} y + 1)

y^{2} - 23 y - 4 = 0

We seek the area of the quadrilateral

O M F N

, where

O (0, 0)

and

F (22, 0)

are vertices on the x-axis. The area is calculated as:

Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times ∣ O F ∣ \times ∣ y_{1} - y_{2} ∣

From the quadratic equation

y^{2} - 23 y - 4 = 0

, we use the properties of roots:

∣ y_{1} - y_{2} ∣ = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2}

∣ y_{1} - y_{2} ∣ = (23)^{2} - 4 (- 4) = 12 + 16 = 28 = 27

Substituting the base

∣ O F ∣ = 22

and the height

27

into the area formula:

Area = \frac{1}{2} \times (22) \times (27)

Area = 214

Common Traps & FAQs

Wait, why is the tangency condition

c^{2} = a^{2} m^{2} - b^{2}

and not

c^{2} = a^{2} m^{2} + b^{2}

? I always get confused between the ellipse and the hyperbola!

That is a classic trap! Think of it this way: for an ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

, the tangent condition is

c^{2} = a^{2} m^{2} + b^{2}

. The plus sign makes sense because the ellipse is 'contained' within a box. For a hyperbola, the geometry is 'open.' The negative sign in the hyperbola equation

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

(which is

\frac{x ^{2}}{4} - \frac{y ^{2}}{4} = 1

) carries over into the tangency condition. Always remember: Ellipse is plus, Hyperbola is minus. Keep that distinction sharp!

Why can't we just find the coordinates of

M

and

N

by solving the quadratic equation? It seems like more work to use the sum and product of roots.

I love that you are looking for the most direct path! While you *could* solve for

y_{1}

and

y_{2}

using the quadratic formula, you will end up with messy radicals. Using the properties of roots (

y_{1} + y_{2}

and

y_{1} y_{2}

) is the 'JEE Advanced way.' It is elegant, faster, and significantly reduces the chance of a silly arithmetic error. In competitive exams, the method that avoids unnecessary calculation is always the superior one.

I am confused about the area calculation. How can we use

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

for a quadrilateral? Isn't that only for triangles?

Great observation! You are right that the standard formula is for triangles. However, look at the quadrilateral

O M F N

. The diagonal

O F

lies entirely on the x-axis. This splits the quadrilateral into two triangles:

△ O F M

and

△ O F N

. Both triangles share the same base

O F

and have heights

∣ y_{1} ∣

and

∣ y_{2} ∣

respectively. The total area is

\frac{1}{2} \times O F \times ∣ y_{1} ∣ + \frac{1}{2} \times O F \times ∣ y_{2} ∣

, which simplifies to

\frac{1}{2} \times O F \times (∣ y_{1} ∣ + ∣ y_{2} ∣)

. Since

M

and

N

are on opposite sides of the x-axis (as the line passes through the focus),

∣ y_{1} ∣ + ∣ y_{2} ∣

is simply the vertical distance

∣ y_{1} - y_{2} ∣

. It is a beautiful geometric shortcut!

Select Answer:

Visualized Solution

Identify the Parabola $P$

Equation of Focal Chord $L$

Hyperbola $H$ Parameters

Tangency Condition

Solve for Slope $m$

Focus $F$ of Hyperbola

Intersection Points $M$ and $N$

Quadratic in $y$

Difference of y-coordinates

Area of Quadrilateral $O M F N$

Final Calculation

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

Analyzing the Setup

The Dance of Tangency

Intersection and Area Calculation

Similar Questions

Let $P (4, 43)$ be a point on the parabola $y^{2} = 4 a x$ and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:

Comprehension Passage

The orthocentre of the triangle $F_{1} M N$ is

If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the x-axis at $Q$ , then the ratio of area of the triangle $M QR$ to area of the quadrilateral $M F_{1} N F_{2}$ is

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

From a point $A$ common tangents are drawn to the circle $x^{2} + y^{2} = a^{2} /2$ and parabola $y^{2} = 4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

Let $A (4, - 4)$ and $B (9, 6)$ be points on the parabola, $y^{2} = 4 x$ . Let $C$ be chosen on the arc $A O B$ of the parabola, where $O$ is the origin, such that the area of $Δ A C B$ is maximum. Then, the area (in sq. units) of $Δ A C B$ , is:

Consider a branch of the hyperbola $x^{2} - 2 y^{2} - 22 x - 42 y - 6 = 0$ with vertex at the point $A$ . Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$ , then the area of the triangle $A B C$ is

Let $O$ be the vertex of the parabola $x^{2} = 4 y$ and $Q$ be any point on it. Let the locus of the point $P$ , which divides the line segment $O Q$ internally in the ratio $2 : 3$ be the conic $C$ . Then the equation of the chord of $C$ , which is bisected at the point $(1, 2)$ , is :

The common tangents to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$ . Then the area of the quadrilateral $P QR S$ is

Select Answer:

Visualized Solution

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

Equation of Focal Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H Parameters

Tangency Condition

Solve for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Focus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F of Hyperbola

Intersection Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Difference of y-coordinates

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

Final Calculation

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

Analyzing the Setup

The Dance of Tangency

Intersection and Area Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocentre of the triangle F1​MN is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

Identify the Parabola $P$

Equation of Focal Chord $L$

Hyperbola $H$ Parameters

Solve for Slope $m$

Focus $F$ of Hyperbola

Intersection Points $M$ and $N$

Quadratic in $y$

Area of Quadrilateral $O M F N$

The orthocentre of the triangle $F_{1} M N$ is

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is