JEE Advanced 1997

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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.

Q: Wait, why do we treat the tangent to $E_1$ as the chord of contact for $E_2$?

That is the core of the problem! The tangent to $E_1$ is a line that intersects $E_2$ at two points, $P$ and $Q$. By definition, the chord of contact of an ellipse with respect to an external point $R$ is the line passing through the points of contact of the tangents drawn from $R$. Since the tangents at $P$ and $Q$ to $E_2$ meet at $R$, the line $PQ$ must be the chord of contact for $E_2$ with respect to $R$. It is a beautiful geometric bridge!

Q: How do we know the locus of $R$ is the Director Circle?

Great observation! After we find the coordinates of $R$ as $(3 \cos \theta, 3 \sin \theta)$, we eliminate $\theta$ to get $x^2 + y^2 = 9$. Now, look at the ellipse $E_2: \frac{x^2}{6} + \frac{y^2}{3} = 1$. The equation of the director circle for any ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2 + b^2$. Here, $a^2 = 6$ and $b^2 = 3$, so $a^2 + b^2 = 9$. Since our locus matches this equation perfectly, $R$ must lie on the director circle.

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the tangent line to the inner ellipse

E_{1}

as it slices through the outer ellipse

E_{2}

. This line segment

P Q

is not just any line; it is the 'Chord of Contact' for the ellipse

E_{2}

with respect to the intersection point

R

of the tangents at

P

and

Q

Hint 2

Hint 2 (The Arsenal): Recall the algebraic condition for the chord of contact of an ellipse

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

from an external point

(x_{1}, y_{1})

, which is

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

. Compare this with the equation of the tangent to

E_{1}

Hint 3

Hint 3 (The Execution): Express the tangent to

E_{1}

in the form

A x + B y = 1

and the chord of contact for

E_{2}

in the form

C x + D y = 1

. Since they are the same line, equate the ratios of the coefficients:

\frac{A}{C} = \frac{B}{D} = \frac{1}{1}

. This will give you the coordinates of

R

in terms of

θ

Visualized Solution

Visualizing the Ellipses $E_{1}$ and $E_{2}$

Given Ellipse $E_{1} : x^{2} + 4 y^{2} = 4$
Given Ellipse $E_{2} : x^{2} + 2 y^{2} = 6$
We need to prove that tangents to $E_{2}$ at the intersection points of a tangent to $E_{1}$ are perpendicular.

Standard Form of Ellipse $E_{1}$

Divide $x^{2} + 4 y^{2} = 4$ by $4$ to get standard form:
$\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$
Here, $a^{2} = 4$ and $b^{2} = 1$
Parametric point on $E_{1}$ : $(2 cos θ, sin θ)$

Equation of Tangent to $E_{1}$

Equation of tangent at $(x_{1}, y_{1})$ is $\frac{x x _{1}}{a ^{2}} + \frac{y y _{1}}{b ^{2}} = 1$
Substituting $(2 cos θ, sin θ)$ :
$\frac{x ( 2 c o s θ )}{4} + \frac{y ( s i n θ )}{1} = 1$
Simplifying: $\frac{x c o s θ}{2} + y sin θ = 1$

Intersection Points $P$ and $Q$

The tangent line $\frac{x c o s θ}{2} + y sin θ = 1$ meets $E_{2}$ at points $P$ and $Q$ .
We are interested in the tangents to $E_{2}$ at these points $P$ and $Q$ .

Intersection Point $R$ of Tangents

Let the tangents at $P$ and $Q$ to $E_{2}$ intersect at $R (x_{1}, y_{1})$ .
Then, the line $P Q$ is the Chord of Contact of $E_{2}$ with respect to point $R$ .

Chord of Contact for $E_{2}$

Standard form of $E_{2} : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$
Equation of chord of contact from $R (x_{1}, y_{1})$ to $E_{2}$ :
$\frac{x x _{1}}{6} + \frac{y y _{1}}{3} = 1$

Identifying the Common Line

Both equations represent the same line $P Q$ :
1) $\frac{c o s θ}{2} x + (sin θ) y = 1$
2) $\frac{x _{1}}{6} x + \frac{y _{1}}{3} y = 1$
Comparing coefficients: $\frac{x _{1} /6}{c o s θ /2} = \frac{y _{1} /3}{s i n θ} = \frac{1}{1}$

Comparing the $x$ Coefficients

From the $x$ ratio: $\frac{x _{1}}{6} = \frac{c o s θ}{2}$
Multiplying by $6$ : $x_{1} = 3 cos θ$

Comparing the $y$ Coefficients

From the $y$ ratio: $\frac{y _{1}}{3} = sin θ$
Multiplying by $3$ : $y_{1} = 3 sin θ$

Finding the Locus of $R$

To find the locus of $R (x_{1}, y_{1})$ , eliminate $θ$ :
$x_{1}^{2} + y_{1}^{2} = (3 cos θ)^{2} + (3 sin θ)^{2}$
$x_{1}^{2} + y_{1}^{2} = 9 (cos^{2} θ + sin^{2} θ)$
$x_{1}^{2} + y_{1}^{2} = 9$

The Director Circle Condition

For $E_{2} : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ , $a^{2} = 6$ and $b^{2} = 3$ .
The equation of its Director Circle is $x^{2} + y^{2} = a^{2} + b^{2}$ .
$x^{2} + y^{2} = 6 + 3 = 9$
The locus of $R$ is the director circle of $E_{2}$ .

Final Proof of Perpendicularity

By definition, the director circle is the locus of points from which perpendicular tangents can be drawn to the ellipse.
Since $R$ lies on the director circle of $E_{2}$ , the tangents $R P$ and $R Q$ are at right angles.
Hence Proved.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler in the realm of coordinate geometry! Today, we explore a problem that feels like a dance between two ellipses.

We are given two ellipses:

E_{1} : x^{2} + 4 y^{2} = 4

E_{2} : x^{2} + 2 y^{2} = 6

We are tasked with proving that if we draw a tangent to the inner ellipse

E_{1}

, it will intersect the outer ellipse

E_{2}

at two points,

P

and

Q

, such that the tangents to

E_{2}

at these points are perpendicular. This is a testament to the hidden symmetry of conic sections.

The Parametric Foundation

First, let us bring our ellipses into their standard forms. For

E_{1}

, dividing by

4

gives:

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

This is an ellipse with

a^{2} = 4

and

b^{2} = 1

. Any point on this ellipse can be elegantly represented using the parameter

θ

(2 cos θ, sin θ)

Using the standard formula for a tangent at

(x_{1}, y_{1})

, which is

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

, we substitute our point to get:

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

Simplifying this, we find the equation of our tangent line:

\frac{x cos θ}{2} + y sin θ = 1

The Bridge of the Chord of Contact

Now, imagine this tangent line extending until it pierces the outer ellipse

E_{2}

at two points,

P

and

Q

. We assume the tangents to

E_{2}

at these points meet at some external point

R (x_{1}, y_{1})

The line

P Q

is the chord of contact for

E_{2}

with respect to the point

R

. The equation for the chord of contact for

E_{2} : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

from point

R (x_{1}, y_{1})

is:

\frac{x x _{1}}{6} + \frac{y y _{1}}{3} = 1

We now have two equations representing the same line

P Q

: the tangent to

E_{1}

and the chord of contact for

E_{2}

The Locus of $R$

Since both equations represent the same line, their coefficients must be proportional. We compare the coefficients:

\frac{x _{1} /6}{cos θ /2} = \frac{y _{1} /3}{sin θ} = \frac{1}{1}

From this, we extract the coordinates of

R

x_{1} = 3 cos θ, y_{1} = 3 sin θ

To find the locus of

R

, we eliminate

θ

by squaring and adding:

x_{1}^{2} + y_{1}^{2} = 9 (cos^{2} θ + sin^{2} θ) = 9

Thus, the locus of

R

is the circle

x^{2} + y^{2} = 9

The Grand Finale

Finally, we look at the properties of

E_{2}

. For

E_{2}

a^{2} = 6

and

b^{2} = 3

The Director Circle of an ellipse is defined as the locus of points from which perpendicular tangents can be drawn, and its equation is

x^{2} + y^{2} = a^{2} + b^{2}

. Substituting our values, we get:

x^{2} + y^{2} = 6 + 3 = 9

Our locus for

R

is exactly the Director Circle of

E_{2}

. Because

R

lies on the Director Circle, the tangents

R P

and

R Q

must be at right angles.

We have successfully navigated the geometry and proven the result.

Common Traps & FAQs

Wait, why do we treat the tangent to

E_{1}

as the chord of contact for

E_{2}

That is the core of the problem! The tangent to

E_{1}

is a line that intersects

E_{2}

at two points,

P

and

Q

. By definition, the chord of contact of an ellipse with respect to an external point

R

is the line passing through the points of contact of the tangents drawn from

R

. Since the tangents at

P

and

Q

E_{2}

meet at

R

, the line

P Q

must be the chord of contact for

E_{2}

with respect to

R

. It is a beautiful geometric bridge!

How do we know the locus of

R

is the Director Circle?

Great observation! After we find the coordinates of

R

(3 cos θ, 3 sin θ)

, we eliminate

θ

to get

x^{2} + y^{2} = 9

. Now, look at the ellipse

E_{2} : \frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1

. The equation of the director circle for any ellipse

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

x^{2} + y^{2} = a^{2} + b^{2}

. Here,

a^{2} = 6

and

b^{2} = 3

, so

a^{2} + b^{2} = 9

. Since our locus matches this equation perfectly,

R

must lie on the director circle.

Is there a simpler way to prove this without using the Director Circle?

While the Director Circle is the most elegant path, you could also calculate the slopes of the tangents

R P

and

R Q

using the coordinates of

P

and

Q

and the point

R

. If you find that the product of their slopes is

- 1

, you have proven they are perpendicular. However, the Director Circle property is a powerful theorem in coordinate geometry that saves you from tedious slope calculations!

Animated Solution for Mathematics - Conic Sections: 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.

Visualized Solution

Visualizing the Ellipses $E_{1}$ and $E_{2}$

Standard Form of Ellipse $E_{1}$

Equation of Tangent to $E_{1}$

Intersection Points $P$ and $Q$

Intersection Point $R$ of Tangents

Chord of Contact for $E_{2}$

Identifying the Common Line

Comparing the $x$ Coefficients

Comparing the $y$ Coefficients

Finding the Locus of $R$

The Director Circle Condition

Final Proof of Perpendicularity

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Foundation

The Bridge of the Chord of Contact

The Locus of $R$

The Grand Finale

Similar Questions

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 :

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 :

If two tangents drawn from a point $(α, β)$ lying on the ellipse $25 x^{2} + 4 y^{2} = 1$ to the parabola $y^{2} = 4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 α + 5)^{2} + (16 β^{2} + 50)^{2}$ equals ____.

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

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

Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on the corresponding directrix.

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:

Visualized Solution

Visualizing the Ellipses .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Standard Form of Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Equation of Tangent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Intersection 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

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

Chord of Contact for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Identifying the Common Line

Comparing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x Coefficients

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

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

The Director Circle Condition

Final Proof of Perpendicularity

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Foundation

The Bridge of the Chord of Contact

The Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

The Grand Finale

Similar Questions

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the parabola y2=4ax is 45∘. Show that the locus of the point P is a hyperbola.

Comprehension Passage

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

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

.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

Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on the corresponding directrix.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the ellipse x2+4y2=4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

Visualizing the Ellipses $E_{1}$ and $E_{2}$

Standard Form of Ellipse $E_{1}$

Equation of Tangent to $E_{1}$

Intersection Points $P$ and $Q$

Intersection Point $R$ of Tangents

Chord of Contact for $E_{2}$

Comparing the $x$ Coefficients

Comparing the $y$ Coefficients

Finding the Locus of $R$

The Locus of $R$

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 :

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

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 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: