JEE Advanced 2012

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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

Q: How do we know the quadratic factor $t^2 + t + 4$ has no real roots?

Always check the discriminant, $D = b^2 - 4ac$. Here, $D = 1^2 - 4(1)(4) = 1 - 16 = -15$. Since the discriminant is negative, the roots are complex. In coordinate geometry problems like this, we are only interested in real intersection points, so we can safely discard this factor.

Enter Numerical Value:

Progressive Hints

Hint 1

The geometry is your compass. Before diving into algebra, visualize the intersection. Both curves share the origin

(0, 0)

. This is your anchor point. The parabola

y^{2} = 8 x

opens right, and the circle

x^{2} + y^{2} - 2 x - 4 y = 0

is centered at

(1, 2)

. Don't just solve; see the shapes.

Hint 2

The algebra of intersection can be messy if you use Cartesian coordinates directly. Instead, embrace the power of parametric coordinates. For the parabola

y^{2} = 8 x

, let

x = 2 t^{2}

and

y = 4 t

. This reduces your system to a single variable

t

, turning a daunting intersection problem into a simple polynomial root-finding exercise.

Hint 3

Once you find the intersection points

P

and

Q

, plot them mentally. You will find that

S (2, 0)

and

Q (2, 4)

share the same x-coordinate. This is the 'Aha!' moment. The segment

S Q

is vertical. If

P S

is horizontal, you have a right-angled triangle. The area is simply

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

Visualized Solution

Visualizing the Setup

Given Parabola: $y^{2} = 8 x$
Given Circle: $x^{2} + y^{2} - 2 x - 4 y = 0$
Objective: Find the area of $Δ P QS$ , where $S$ is the focus and $P Q$ is the common chord.

Finding the Focus $S$

Compare $y^{2} = 8 x$ with standard form $y^{2} = 4 a x$
Here, $4 a = 8 \Rightarrow a = 2$
Focus $S = (a, 0) = (2, 0)$

Visualizing the Curves

Parabola: $y^{2} = 8 x$
Circle: $x^{2} + y^{2} - 2 x - 4 y = 0$
Observe that both curves pass through the origin $(0, 0)$ .

Using Parametric Coordinates

For $y^{2} = 8 x$ , parameter $a = 2$
Parametric point on parabola: $Q = (2 t^{2}, 4 t)$
We will substitute these coordinates into the circle's equation.

The Substitution Step

Substitute $x = 2 t^{2}$ and $y = 4 t$ into $x^{2} + y^{2} - 2 x - 4 y = 0$ :
$(2 t^{2})^{2} + (4 t)^{2} - 2 (2 t^{2}) - 4 (4 t) = 0$

Algebraic Simplification

Expand the terms:
$4 t^{4} + 16 t^{2} - 4 t^{2} - 16 t = 0$
Combine like terms:
$4 t^{4} + 12 t^{2} - 16 t = 0$

Solving for Parameter $t$

Divide by $4$ : $t^{4} + 3 t^{2} - 4 t = 0$
Factor out $t$ : $t (t^{3} + 3 t - 4) = 0$
Factor completely: $t (t - 1) (t^{2} + t + 4) = 0$

Finding Points $P$ and $Q$

Real roots: $t = 0$ and $t = 1$ (since $t^{2} + t + 4 = 0$ has imaginary roots)
For $t = 0 \Rightarrow P = (0, 0)$
For $t = 1 \Rightarrow Q = (2, 4)$

Drawing the Common Chord $P Q$

The common chord $P Q$ connects $P (0, 0)$ and $Q (2, 4)$ .

Analyzing Triangle $P QS$

Vertices of $Δ P QS$ : $P (0, 0)$ , $Q (2, 4)$ , $S (2, 0)$
Observe: $S (2, 0)$ and $Q (2, 4)$ have the same x-coordinate.
Therefore, $S Q$ is a vertical line segment.

Identifying the Right Angle

Segment $P S$ lies along the x-axis.
Segment $S Q$ is vertical (parallel to the y-axis).
Therefore, $∠ P S Q = 9 0^{\circ}$ .

Calculating Base and Height

Base $P S = distance from (0, 0) to (2, 0) = 2$
Height $S Q = distance from (2, 0) to (2, 4) = 4$

Calculating the Final Area

Area of $Δ P QS = \frac{1}{2} \times base \times height$
Area $= \frac{1}{2} \times P S \times S Q$
Area $= \frac{1}{2} \times 2 \times 4 = 4$ sq. units

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

Analyzing the Setup

Welcome, future engineer. Today, we aren't just solving a problem; we are embarking on a journey through the elegant landscape of coordinate geometry.

We have a parabola,

y^{2} = 8 x

, and a circle,

x^{2} + y^{2} - 2 x - 4 y = 0

. They dance together, intersecting at points

P

and

Q

. Our mission is to find the area of the triangle formed by these points and the parabola's focus,

S

The Anchor Point

First, let's find our anchor: the focus

S

. Comparing

y^{2} = 8 x

to the standard form

y^{2} = 4 a x

, we see

4 a = 8

, which implies

a = 2

Our focus

S

sits proudly at

(2, 0)

. Both curves pass through the origin

(0, 0)

, which serves as our first intersection point,

P

The Parametric Leap

To find

Q

, we summon the power of parametric coordinates. By setting

x = 2 t^{2}

and

y = 4 t

, we transform the parabola into a single variable

t

Substituting these into the circle's equation, we get:

(2 t^{2})^{2} + (4 t)^{2} - 2 (2 t^{2}) - 4 (4 t) = 0

Expanding this, we arrive at:

4 t^{4} + 16 t^{2} - 4 t^{2} - 16 t = 0

This simplifies to the polynomial:

4 t^{4} + 12 t^{2} - 16 t = 0

The Algebraic Dance

Dividing by

4

, we obtain

t^{4} + 3 t^{2} - 4 t = 0

. Factoring out

t

, we have:

t (t^{3} + 3 t - 4) = 0

By inspection,

t = 1

is a root. Factoring completely, we find:

t (t - 1) (t^{2} + t + 4) = 0

The quadratic factor

t^{2} + t + 4

has a negative discriminant (

D = 1^{2} - 4 (1) (4) = - 15

), so it yields no real roots. Thus, our real intersection points occur at

t = 0

and

t = 1

For

t = 0

, we get

P (0, 0)

. For

t = 1

, we get

Q (2, 4)

The Geometric Revelation

Now, consider the triangle

P QS

with vertices

P (0, 0)

Q (2, 4)

, and

S (2, 0)

. Notice that

S

and

Q

share the same x-coordinate of

2

This means

S Q

is a vertical line segment. Since

P S

lies on the x-axis, the angle at

S

9 0^{\circ}

. We have a right-angled triangle!

The base

P S

is the distance from

(0, 0)

(2, 0)

, which is

2

. The height

S Q

is the vertical distance from

(2, 0)

(2, 4)

, which is

4

The area is calculated as:

Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 4 = 4

The final area of the triangle is 4 square units.

Common Traps & FAQs

Why did we use parametric coordinates instead of solving

y^{2} = 8 x

for

x

and substituting into the circle?

Excellent question. While substitution works, it leads to a quartic equation in

y

that is often harder to factor. Parametric coordinates

x = 2 t^{2}, y = 4 t

map the parabola perfectly and reduce the degree of the equation, making the roots

t = 0

and

t = 1

pop out almost instantly. It is about choosing the path of least resistance.

How do we know the quadratic factor

t^{2} + t + 4

has no real roots?

Always check the discriminant,

D = b^{2} - 4 a c

. Here,

D = 1^{2} - 4 (1) (4) = 1 - 16 = - 15

. Since the discriminant is negative, the roots are complex. In coordinate geometry problems like this, we are only interested in real intersection points, so we can safely discard this factor.

Is there a way to solve this without finding the coordinates of

Q

Technically, you could use the formula for the area of a triangle formed by the intersection of a line and a curve, but that is overkill. In JEE Advanced, the most elegant solution is often the one that reveals the geometry. Finding the coordinates

P

and

Q

is fast and gives you the exact vertices, which makes calculating the area trivial.

Animated Solution for Mathematics - Conic Sections: 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

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Finding the Focus $S$

Visualizing the Curves

Using Parametric Coordinates

The Substitution Step

Algebraic Simplification

Solving for Parameter $t$

Finding Points $P$ and $Q$

Drawing the Common Chord $P Q$

Analyzing Triangle $P QS$

Identifying the Right Angle

Calculating Base and Height

Calculating the Final Area

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

Analyzing the Setup

The Anchor Point

The Parametric Leap

The Algebraic Dance

The Geometric Revelation

Similar Questions

Tangents drawn from the point (-8, 0) to the parabola $y^{2} = 8 x$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :-

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

Let $A$ be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ intersect the parabola at two distinct points $B$ and $C$ . If the centroid of the triangle $A B C$ is $(\frac{7}{3}, \frac{4}{3})$ , then $(B C)^{2}$ is equal to :

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:

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.

Comprehension Passage

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is

A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^{2} = 2 x$ and touches the parabola $y = (x - \frac{1}{4})^{2} + α$ , where $α > 0$ . Then $(4 α - 8)^{2}$ is equal to ____.

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $2$ and $22$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

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

Visualizing the Curves

Using Parametric Coordinates

The Substitution Step

Algebraic Simplification

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

Finding Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Drawing the Common Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Analyzing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQS

Identifying the Right Angle

Calculating Base and Height

Calculating the Final Area

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

Analyzing the Setup

The Anchor Point

The Parametric Leap

The Algebraic Dance

The Geometric Revelation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of the areas of the triangles PQS and PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circumcircle of the triangle PRS is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the incircle of the triangle PQR is

Finding the Focus $S$

Solving for Parameter $t$

Finding Points $P$ and $Q$

Drawing the Common Chord $P Q$

Analyzing Triangle $P QS$

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is