JEE Main 2021 (25 July Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: If a tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the tangents at the extremities of its major axis at $B$ and $C$ , then the circle with $B C$ as diameter passes through the point:

Q: Wait, why do we use parametric coordinates instead of the standard $y = mx + c$ form for the tangent?

That is a fantastic question! While you *could* use $y = mx + c$, the condition for tangency $c^2 = a^2m^2 + b^2$ introduces a square root and makes the algebra significantly more cumbersome. By using parametric coordinates $P(2\cos\theta, \sin\theta)$, we reduce the entire problem to a single variable $\theta$. It turns the messy algebra into a beautiful trigonometric dance where identities like $\tan\frac{\theta}{2} + \cot\frac{\theta}{2} = \frac{2}{\sin\theta}$ simplify the expression almost magically. Always look for the path of least algebraic resistance!

Q: Why are the tangents at the extremities of the major axis vertical lines?

Think about the orientation of the ellipse. The equation $\frac{x^2}{4} + \frac{y^2}{1} = 1$ tells us the major axis lies along the x-axis. The extremities of the major axis are the vertices $(2, 0)$ and $(-2, 0)$. At these points, the tangent is perpendicular to the major axis. Since the major axis is horizontal, the tangent must be vertical. A vertical line passing through $(2, 0)$ is simply $x = 2$, and the one through $(-2, 0)$ is $x = -2$. It is a simple geometric consequence of the ellipse's symmetry!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. You have a fixed ellipse and a tangent line that slides along it. The points

B

and

C

are the intersections of this moving tangent with the two fixed vertical lines

x = 2

and

x = - 2

. The circle is defined by the diameter

B C

. Think about how the position of the tangent changes as the point of tangency moves, and ask yourself: what geometric property must remain invariant for the circle to always pass through a specific point?

Hint 2

Hint 2 (The Arsenal): You need to parameterize the ellipse. Use the parametric coordinates

P (2 cos θ, sin θ)

. Recall the equation of a tangent to the ellipse

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

at point

(a cos θ, b sin θ)

, which is

\frac{x c o s θ}{a} + \frac{y s i n θ}{b} = 1

. Also, remember the diameter form of a circle equation:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

, where

(x_{1}, y_{1})

and

(x_{2}, y_{2})

are the endpoints of the diameter.

Hint 3

Hint 3 (The Execution): Start by finding the coordinates of

B

and

C

in terms of

θ

. Substitute

x = 2

and

x = - 2

into the tangent equation

\frac{x c o s θ}{2} + y sin θ = 1

. Once you have

B (- 2, cot \frac{θ}{2})

and

C (2, tan \frac{θ}{2})

, plug these into the circle equation

(x - 2) (x + 2) + (y - tan \frac{θ}{2}) (y - cot \frac{θ}{2}) = 0

. Simplify this expression, and look for the term that depends on

θ

—you will need to make that term vanish to find the fixed point.

Visualized Solution

Standard Form of the Ellipse

Given equation: $x^{2} + 4 y^{2} = 4$
Divide by $4$ to get standard form: $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$
Comparing with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , we get $a = 2, b = 1$

Tangents at Major Axis Extremities

Extremities of the major axis are at $x = a$ and $x = - a$ .
Here, the points are $A (2, 0)$ and $A^{'} (- 2, 0)$ .
The tangents at these extremities are vertical lines: $x = 2$ and $x = - 2$ .

Equation of the General Tangent

Let a general point $P$ on the ellipse be $(2 cos θ, sin θ)$ .
The equation of the tangent at $P$ is $\frac{x ( 2 c o s θ )}{4} + \frac{y ( s i n θ )}{1} = 1$ .
Simplifying, we get: $\frac{x c o s θ}{2} + y sin θ = 1$ .

Finding Intersection Point $C$

The general tangent meets $x = 2$ at point $C$ .
Substitute $x = 2$ into $\frac{x c o s θ}{2} + y sin θ = 1$ .
$cos θ + y sin θ = 1 ⟹ y = \frac{1 - c o s θ}{s i n θ}$ .
Using half-angle formulas, $y = tan \frac{θ}{2}$ . So, $C (2, tan \frac{θ}{2})$ .

Finding Intersection Point $B$

The general tangent meets $x = - 2$ at point $B$ .
Substitute $x = - 2$ into $\frac{x c o s θ}{2} + y sin θ = 1$ .
$- cos θ + y sin θ = 1 ⟹ y = \frac{1 + c o s θ}{s i n θ}$ .
Using half-angle formulas, $y = cot \frac{θ}{2}$ . So, $B (- 2, cot \frac{θ}{2})$ .

Equation of the Circle with Diameter $B C$

We need the circle with $B C$ as its diameter.
Using the diameter form: $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .
Substitute $B (- 2, cot \frac{θ}{2})$ and $C (2, tan \frac{θ}{2})$ .
$(x + 2) (x - 2) + (y - cot \frac{θ}{2}) (y - tan \frac{θ}{2}) = 0$ .

Simplifying the Circle Equation

Expand the $x$ terms: $(x + 2) (x - 2) = x^{2} - 4$ .
Expand the $y$ terms: $y^{2} - y (tan \frac{θ}{2} + cot \frac{θ}{2}) + tan \frac{θ}{2} cot \frac{θ}{2}$ .
Note that $tan \frac{θ}{2} cot \frac{θ}{2} = 1$ .
Also, $tan \frac{θ}{2} + cot \frac{θ}{2} = \frac{1}{s i n \frac{θ}{2} c o s \frac{θ}{2}} = \frac{2}{s i n θ}$ .
The circle equation becomes: $x^{2} + y^{2} - \frac{2 y}{s i n θ} - 3 = 0$ .

Finding the Fixed Point

The circle equation is $x^{2} + y^{2} - 3 - \frac{2 y}{s i n θ} = 0$ .
This represents a family of circles. For the circle to pass through a fixed point independent of $θ$ , the coefficient of the variable part must be zero.
Set $y = 0$ to eliminate the $θ$ dependence.
Substitute $y = 0$ into the equation: $x^{2} + 0 - 3 - 0 = 0 ⟹ x^{2} = 3$ .
Therefore, $x = \pm 3$ . The fixed points are $(3, 0)$ and $(- 3, 0)$ .

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The ellipse is defined by the equation

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

. By dividing the entire equation by

4

, we obtain the standard form:

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

Here, we identify the semi-major axis

a = 2

and the semi-minor axis

b = 1

. The vertices of the ellipse are located at

A (2, 0)

and

A^{'} (- 2, 0)

The Vertical Guardians

At these vertices, we draw two vertical tangents. These lines act as boundaries at

x = 2

and

x = - 2

A third tangent line slides along the curve of the ellipse. As it moves, it intersects these two vertical lines at points

B

and

C

. Our goal is to find the fixed points through which the circle with diameter

B C

always passes.

The Parametric Elegance

To describe the motion of the sliding tangent, we use parametric coordinates. Any point

P

on the ellipse can be represented as

(2 cos θ, sin θ)

The equation of the tangent at point

P

is given by:

\frac{x ( 2 cos θ )}{4} + \frac{y ( sin θ )}{1} = 1

This simplifies to the following linear equation:

\frac{x cos θ}{2} + y sin θ = 1

The Intersection and the Half-Angle Magic

To find point

C

, we set

x = 2

in the tangent equation:

\frac{2 cos θ}{2} + y sin θ = 1 \Rightarrow cos θ + y sin θ = 1

Solving for

y

, we get

y = \frac{1 - c o s θ}{s i n θ}

. Using the trigonometric identity

\frac{1 - c o s θ}{s i n θ} = tan (\frac{θ}{2})

, we find that point

C

is at:

C = (2, tan \frac{θ}{2})

By symmetry, setting

x = - 2

to find point

B

yields

- cos θ + y sin θ = 1

, which simplifies to

y = \frac{1 + c o s θ}{s i n θ} = cot (\frac{θ}{2})

. Thus, point

B

is at:

B = (- 2, cot \frac{θ}{2})

The Circle's Identity

We construct the circle using the diameter form

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

. Substituting the coordinates of

B

and

C

(x + 2) (x - 2) + (y - cot \frac{θ}{2}) (y - tan \frac{θ}{2}) = 0

Expanding this expression, we get:

x^{2} - 4 + y^{2} - y (tan \frac{θ}{2} + cot \frac{θ}{2}) + tan \frac{θ}{2} cot \frac{θ}{2} = 0

Since

tan (\frac{θ}{2}) cot (\frac{θ}{2}) = 1

, the constant terms combine to

- 3

. Furthermore,

tan (\frac{θ}{2}) + cot (\frac{θ}{2}) = \frac{2}{s i n θ}

. The equation of the circle becomes:

x^{2} + y^{2} - 3 - \frac{2 y}{sin θ} = 0

The Revelation

This equation represents a family of circles. For the circle to pass through a fixed point independent of

θ

, the term containing

θ

must vanish. This occurs when

y = 0

Substituting

y = 0

into the equation, we are left with:

x^{2} - 3 = 0 \Rightarrow x = \pm 3

Thus, the circle always passes through the fixed points $(3, 0)$ and $(- 3, 0)$ .

Common Traps & FAQs

Wait, why do we use parametric coordinates instead of the standard

y = m x + c

form for the tangent?

That is a fantastic question! While you *could* use

y = m x + c

, the condition for tangency

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

introduces a square root and makes the algebra significantly more cumbersome. By using parametric coordinates

P (2 cos θ, sin θ)

, we reduce the entire problem to a single variable

θ

. It turns the messy algebra into a beautiful trigonometric dance where identities like

tan \frac{θ}{2} + cot \frac{θ}{2} = \frac{2}{s i n θ}

simplify the expression almost magically. Always look for the path of least algebraic resistance!

How do we know the circle passes through a fixed point? The equation still has

θ

in it!

You are absolutely right to be skeptical. The equation we derived,

x^{2} + y^{2} - 3 - \frac{2 y}{s i n θ} = 0

, represents a *family* of circles, one for every value of

θ

. For a point to be 'fixed' (meaning it lies on every circle in this family), its coordinates must satisfy the equation regardless of the value of

θ

. This implies that the term containing

θ

must vanish. Since the term is

- \frac{2 y}{s i n θ}

, the only way for this to be zero for all

θ

is if

y = 0

. Setting

y = 0

forces the

θ

-dependent part to disappear, leaving us with a constant equation for

x

Why are the tangents at the extremities of the major axis vertical lines?

Think about the orientation of the ellipse. The equation

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

tells us the major axis lies along the x-axis. The extremities of the major axis are the vertices

(2, 0)

and

(- 2, 0)

. At these points, the tangent is perpendicular to the major axis. Since the major axis is horizontal, the tangent must be vertical. A vertical line passing through

(2, 0)

is simply

x = 2

, and the one through

(- 2, 0)

x = - 2

. It is a simple geometric consequence of the ellipse's symmetry!

Animated Solution for Mathematics - Conic Sections: If a tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the tangents at the extremities of its major axis at $B$ and $C$ , then the circle with $B C$ as diameter passes through the point:

Select Answer:

Visualized Solution

Standard Form of the Ellipse

Tangents at Major Axis Extremities

Equation of the General Tangent

Finding Intersection Point $C$

Finding Intersection Point $B$

Equation of the Circle with Diameter $B C$

Simplifying the Circle Equation

Finding the Fixed Point

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Vertical Guardians

The Parametric Elegance

The Intersection and the Half-Angle Magic

The Circle's Identity

The Revelation

Similar Questions

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

On the ellipse $4 x^{2} + 9 y^{2} = 1$ , the points at which the tangents are parallel to the line $8 x = 9 y$ are

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

Let $T_{1}, T_{2}$ be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :-

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

A tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the ellipse $x^{2} + 2 y^{2} = 6$ at $P$ and $Q$ . Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2} + 2 y^{2} = 6$ are at right angles.

Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

Select Answer:

Visualized Solution

Standard Form of the Ellipse

Tangents at Major Axis Extremities

Equation of the General Tangent

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Equation of the Circle with Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Simplifying the Circle Equation

Finding the Fixed Point

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Vertical Guardians

The Parametric Elegance

The Intersection and the Half-Angle Magic

The Circle's Identity

The Revelation

Similar Questions

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Comprehension Passage

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the locus of the point whose distances from the point P and the line AB are equal, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

Finding Intersection Point $C$

Finding Intersection Point $B$

Equation of the Circle with Diameter $B C$

On the ellipse $4 x^{2} + 9 y^{2} = 1$ , the points at which the tangents are parallel to the line $8 x = 9 y$ are

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :-

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

A tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the ellipse $x^{2} + 2 y^{2} = 6$ at $P$ and $Q$ . Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2} + 2 y^{2} = 6$ are at right angles.

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The orthocenter of the triangle $P A B$ is

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If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :