JEE Advanced 2009

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Match the conics in Column I with the statements/expressions in Column II.

List-I

(P)

Circle

(Q)

Parabola

(R)

Ellipse

(S)

Hyperbola

List-II

(1)

The locus of the point

(h, k)

for which the line

h x + k y = 1

touches the circle

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

(2)

Points

z

in the complex plane satisfying

∣ z + 2∣ - ∣ z - 2∣ = \pm 3

(3)

Points of the conic have parametric representation

x = 3 (\frac{1 - t ^{2}}{1 + t ^{2}}), y = \frac{2 t}{1 + t ^{2}}

(4)

The eccentricity of the conic lies in the interval

1 \leq x < \infty

(5)

Points

z

in the complex plane satisfying

R e (z + 1)^{2} = ∣ z ∣^{2} + 1

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Visualized Solution

Introduction to Conic Matching

Objective: Match conics in Column I with their properties in Column II.
Column I: (A) Circle, (B) Parabola, (C) Ellipse, (D) Hyperbola.
Column II: Locus problems, complex equations, and parametric forms.

Statement (p): Tangency Condition

Line $h x + k y = 1$ touches the circle $x^{2} + y^{2} = 4$ .
Center of the circle is $(0, 0)$ and radius is $r = 2$ .
For tangency, the perpendicular distance from the center to the line must equal the radius.

Statement (p): Deriving the Locus

Distance formula: $\frac{∣ h ( 0 ) + k ( 0 ) - 1∣}{h ^{2} + k ^{2}} = 2$
Simplifying: $\frac{1}{h ^{2} + k ^{2}} = 2$
Squaring both sides: $h^{2} + k^{2} = \frac{1}{4}$
The locus of $(h, k)$ is a Circle. Matches (A).

Statement (q): Complex Locus Definition

Equation: $∣ z + 2∣ - ∣ z - 2∣ = \pm 3$
Recall the standard definition: $∣ z - z_{1} ∣ - ∣ z - z_{2} ∣ = \pm 2 a$
This represents a Hyperbola if $2 a < ∣ z_{1} - z_{2} ∣$ .

Statement (q): Validating the Hyperbola

Foci are at $z_{1} = - 2$ and $z_{2} = 2$ .
Distance between foci: $∣ z_{1} - z_{2} ∣ = 4$ .
Constant difference: $2 a = 3$ .
Since $3 < 4$ , the locus is a valid Hyperbola. Matches (D).

Statement (r): Parametric Representation

Given: $x = 3 (\frac{1 - t ^{2}}{1 + t ^{2}})$ and $y = \frac{2 t}{1 + t ^{2}}$
Let's use trigonometric substitution: $t = tan (θ /2)$
This transforms the rational expressions into standard trigonometric functions.

Statement (r): Eliminating the Parameter

$\frac{1 - t ^{2}}{1 + t ^{2}} = cos θ$ and $\frac{2 t}{1 + t ^{2}} = sin θ$
So, $x = 3 cos θ$ and $y = sin θ$
Rearranging: $\frac{x}{3} = cos θ$ and $y = sin θ$
Squaring and adding: $\frac{x ^{2}}{3} + y^{2} = 1$
This is an Ellipse. Matches (C).

Statement (s): The Eccentricity Interval

Given interval for eccentricity $e$ : $1 \leq e < \infty$
For a Parabola, eccentricity is exactly $e = 1$ .
For a Hyperbola, eccentricity is strictly $e > 1$ .
Therefore, the interval includes both Parabola and Hyperbola. Matches (B) and (D).

Statement (t): Expanding the Complex Equation

Equation: $R e (z + 1)^{2} = ∣ z ∣^{2} + 1$
Let $z = x + i y$ . Then $z + 1 = (x + 1) + i y$ .
Expand $(z + 1)^{2}$ : $((x + 1) + i y)^{2} = (x + 1)^{2} - y^{2} + 2 i (x + 1) y$

Statement (t): Extracting the Real Part

The real part of $(z + 1)^{2}$ is $(x + 1)^{2} - y^{2}$ .
The right side is $∣ z ∣^{2} + 1 = x^{2} + y^{2} + 1$ .
Equating them: $(x + 1)^{2} - y^{2} = x^{2} + y^{2} + 1$

Statement (t): Finalizing the Parabola Equation

Expand the left side: $x^{2} + 2 x + 1 - y^{2} = x^{2} + y^{2} + 1$
Cancel common terms ( $x^{2}$ and $1$ ) from both sides.
We get: $2 x - y^{2} = y^{2} ⟹ 2 y^{2} = 2 x ⟹ y^{2} = x$
This is the equation of a Parabola. Matches (B).

Final Matching Summary

(A) Circle $\to$ (p)
(B) Parabola $\to$ (s), (t)
(C) Ellipse $\to$ (r)
(D) Hyperbola $\to$ (q), (s)
Key Takeaway: Conics can be represented in multiple ways: loci, complex equations, parametric forms, and eccentricity.

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

Solution Diagram

Analyzing the Circle

When we see the line

h x + k y = 1

touching the circle

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

, we must immediately recall the geometric definition of tangency. The perpendicular distance from the center

(0, 0)

to the line must equal the radius

r = 2

Using the distance formula, we obtain:

\frac{∣ - 1∣}{h ^{2} + k ^{2}} = 2

Squaring this, we find

h^{2} + k^{2} = \frac{1}{4}

. This result confirms that the locus of the point

(h, k)

is a circle.

The Hyperbola in the Complex Plane

Next, we encounter the complex plane. The equation

∣ z + 2∣ - ∣ z - 2∣ = \pm 3

is a classic representation of a conic section.

It represents the difference of distances from two fixed points (foci) being constant. This is the definition of a hyperbola.

We check the condition: the distance between foci is

4

, and the constant difference

2 a

3

. Since

3 < 4

, the hyperbola exists.

Parametric Forms and the Ellipse

Moving to parametric forms, we see

x = 3 (\frac{1 - t ^{2}}{1 + t ^{2}})

and

y = \frac{2 t}{1 + t ^{2}}

. This structure suggests trigonometric substitution.

By setting

t = tan (θ /2)

, we transform these into

x = 3 cos θ

and

y = sin θ

. Squaring and adding, we get:

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

This is the elegant equation of an ellipse.

The Parabola via Complex Algebra

Finally, we tackle the complex equation

R e (z + 1)^{2} = ∣ z ∣^{2} + 1

. By substituting

z = x + i y

, we expand the real part to get

(x + 1)^{2} - y^{2}

Equating this to

x^{2} + y^{2} + 1

, the quadratic terms cancel out:

x^{2} + 2 x + 1 - y^{2} = x^{2} + y^{2} + 1

Simplifying this expression, we are left with

2 x = 2 y^{2}

, or

y^{2} = x

, which is the standard form of a parabola.

Mathematics is not about memorizing formulas; it is about seeing the underlying structure. Keep practicing, and these shapes will become second nature to you.

Common Traps & FAQs

Why is the condition

∣ z_{1} - z_{2} ∣ > 2 a

crucial for the hyperbola definition?

This is the triangle inequality in disguise. If the distance between the foci is less than the constant difference, the locus is empty because no point can satisfy the condition. It is the geometric 'existence' check for the hyperbola.

Why did we use the substitution

t = tan (θ /2)

for the parametric form?

This is the Weierstrass substitution. It is a powerful tool to convert rational functions of

t

into trigonometric functions of

θ

. It is the standard 'key' to unlock parametric forms involving

(1 - t^{2}) / (1 + t^{2})

Can we always just expand

z = x + i y

for complex loci?

Yes, it is the 'brute force' method that always works. While geometric intuition is faster, algebra is your safety net. If you get stuck, convert to Cartesian coordinates and grind through the algebra.

Animated Solution for Mathematics - Conic Sections: Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to Conic Matching

Statement (p): Tangency Condition

Statement (p): Deriving the Locus

Statement (q): Complex Locus Definition

Statement (q): Validating the Hyperbola

Statement (r): Parametric Representation

Statement (r): Eliminating the Parameter

Statement (s): The Eccentricity Interval

Statement (t): Expanding the Complex Equation

Statement (t): Extracting the Real Part

Statement (t): Finalizing the Parabola Equation

Final Matching Summary

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

Analyzing the Circle

The Hyperbola in the Complex Plane

Parametric Forms and the Ellipse

The Parabola via Complex Algebra

Similar Questions

Let $S = {(x, y) \in R^{2} : \frac{y ^{2}}{1 + r} - \frac{x ^{2}}{1 - r} = 1}$ , where $r \neq = \pm 1$ . Then S represents :

The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

A circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle 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 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

The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

The curve described parametrically by $x = t^{2} + t + 1, y = t^{2} - t + 1$ represents

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P (\frac{1}{2}, 1)$ . Its one directrix is the common tangent, nearer to the point $P$ , to the circle $x^{2} + y^{2} = 1$ and the hyperbola $x^{2} - y^{2} = 1$ . The equation of the ellipse, in the standard form, is.........

Animated Solution for Mathematics - Conic Sections: Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to Conic Matching

Statement (p): Tangency Condition

Statement (p): Deriving the Locus

Statement (q): Complex Locus Definition

Statement (q): Validating the Hyperbola

Statement (r): Parametric Representation

Statement (r): Eliminating the Parameter

Statement (s): The Eccentricity Interval

Statement (t): Expanding the Complex Equation

Statement (t): Extracting the Real Part

Statement (t): Finalizing the Parabola Equation

Final Matching Summary

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

Analyzing the Circle

The Hyperbola in the Complex Plane

Parametric Forms and the Ellipse

The Parabola via Complex Algebra

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the point of intersection of the lines (3​)kx+ky−43​=0 and 3​x−y−4(3​)k=0 is a conic, whose eccentricity is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid point of the line segment joining the point (4,3) and the points on the ellipse x2+2y2=4 is an ellipse with eccentricity :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve described parametrically by x=t2+t+1,y=t2−t+1 represents

Let $S = {(x, y) \in R^{2} : \frac{y ^{2}}{1 + r} - \frac{x ^{2}}{1 - r} = 1}$ , where $r \neq = \pm 1$ . Then S represents :

The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

A circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle is

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The curve described parametrically by $x = t^{2} + t + 1, y = t^{2} - t + 1$ represents