Wait, why do we use the eccentric angle $\phi$ instead of the actual angle $\theta$ from the center of the ellipse?

Great question! The eccentric angle $\phi$ is defined specifically for the ellipse using the auxiliary circle. The angle $\theta$ from the center (polar angle) doesn't have the same clean relationship with the normal equation. The auxiliary circle is the 'master key' for ellipse problems.

JEE Advanced 2000

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let $A B C$ be an equilateral triangle inscribed in the circle $x^{2} + y^{2} = a^{2}$ . Suppose perpendiculars from $A, B, C$ to the major axis of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, (a > b)$ meets the ellipse respectively, at $P, Q, R$ , so that $P, Q, R$ lie on the same side of the major axis as $A, B, C$ respectively. Prove that the normals to the ellipse drawn at the points $P, Q$ and $R$ are concurrent.

Visualized Solution

The Auxiliary Circle & Triangle

Consider the auxiliary circle $x^{2} + y^{2} = a^{2}$ .
An equilateral triangle $A B C$ is inscribed in this circle.
Let the eccentric angle of point $A$ be $α$ .

Eccentric Angles of $B$ and $C$

Since $△ A B C$ is equilateral, the points are separated by $\frac{2 π}{3}$ radians ( $12 0^{\circ}$ ).
Eccentric angle of $B$ : $ϕ_{2} = α + \frac{2 π}{3}$
Eccentric angle of $C$ : $ϕ_{3} = α + \frac{4 π}{3}$

Projecting onto the Ellipse

The ellipse equation is $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ .
Perpendiculars are dropped from $A, B, C$ to the major axis.
These perpendiculars meet the ellipse at points $P, Q, R$ .

Eccentric Angles of $P, Q, R$

By definition of the auxiliary circle, points on the same vertical line share the same eccentric angle.
Therefore, $P, Q, R$ have the exact same eccentric angles as $A, B, C$ .
$ϕ_{1} = α$ , $ϕ_{2} = α + \frac{2 π}{3}$ , $ϕ_{3} = α + \frac{4 π}{3}$

Equation of the Normal

The general equation of a normal to the ellipse at eccentric angle $ϕ$ is:
$a x sec ϕ - b y csc ϕ = a^{2} - b^{2}$

Rearranging the Normal Equation

Multiply the entire equation by $sin ϕ cos ϕ$ :
$a x sin ϕ - b y cos ϕ = (a^{2} - b^{2}) sin ϕ cos ϕ$
Using $2 sin ϕ cos ϕ = sin 2 ϕ$ :
$a x sin ϕ - b y cos ϕ = \frac{a ^{2} - b ^{2}}{2} sin 2 ϕ$

Condition for Concurrency of Normals

For three normals at angles $ϕ_{1}, ϕ_{2}, ϕ_{3}$ to be concurrent, their eccentric angles must satisfy:
$sin (ϕ_{1} + ϕ_{2}) + sin (ϕ_{2} + ϕ_{3}) + sin (ϕ_{3} + ϕ_{1}) = 0$

Substituting the Angles

Let's calculate the pairwise sums:
$ϕ_{1} + ϕ_{2} = 2 α + \frac{2 π}{3}$
$ϕ_{2} + ϕ_{3} = 2 α + 2 π$
$ϕ_{3} + ϕ_{1} = 2 α + \frac{4 π}{3}$

Setting up the Sum

We need to evaluate:
$S = sin (2 α + \frac{2 π}{3}) + sin (2 α + 2 π) + sin (2 α + \frac{4 π}{3})$
Note that $sin (2 α + 2 π) = sin (2 α)$
And $\frac{4 π}{3}$ is equivalent to $- \frac{2 π}{3}$ for the sine function's periodicity.

Applying Trigonometric Identities

Rewrite the sum:
$S = sin (2 α) + [sin (2 α + \frac{2 π}{3}) + sin (2 α - \frac{2 π}{3})]$
Use the identity: $sin (A + B) + sin (A - B) = 2 sin A cos B$
Here, $A = 2 α$ and $B = \frac{2 π}{3}$

Final Evaluation

$S = sin (2 α) + 2 sin (2 α) cos (\frac{2 π}{3})$
We know $cos (\frac{2 π}{3}) = - \frac{1}{2}$
$S = sin (2 α) + 2 sin (2 α) (- \frac{1}{2})$
$S = sin (2 α) - sin (2 α) = 0$

Conclusion

The concurrency condition $\sum sin (ϕ_{i} + ϕ_{j}) = 0$ is perfectly satisfied.
Therefore, the normals to the ellipse at points $P, Q,$ and $R$ are concurrent.
Pro Tip: This property holds for any equilateral triangle inscribed in the auxiliary circle!

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on the auxiliary circle of an ellipse, a perfect circle of radius

a

. Inside this circle, we have inscribed an equilateral triangle

A B C

It is a symbol of perfect symmetry. We drop perpendiculars from these vertices to the major axis of our ellipse, defined by the equation:

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

These perpendiculars land on the ellipse at points

P, Q,

and

R

. Our mission is to prove that the normals at these points are concurrent.

The Bridge of Eccentric Angles

First, we identify the eccentric angles. Since

A B C

is an equilateral triangle, the angles are

α

α + \frac{2 π}{3}

, and

α + \frac{4 π}{3}

Because the ellipse is a scaled version of the auxiliary circle, the points

P, Q,

and

R

share these exact same eccentric angles. This is our bridge, serving as the fundamental connection between the circle and the ellipse.

The Algebra of Normals

The standard normal equation

a x sec ϕ - b y csc ϕ = a^{2} - b^{2}

is often cumbersome. Let us transform it by multiplying by

sin ϕ cos ϕ

a x sin ϕ - b y cos ϕ = \frac{a ^{2} - b ^{2}}{2} sin 2 ϕ

This form is much more manageable. For three normals to be concurrent, the condition

\sum sin (ϕ_{i} + ϕ_{j}) = 0

must be satisfied.

Substituting our specific angles, we obtain the expression:

sin (2 α + \frac{2 π}{3}) + sin (2 α + 2 π) + sin (2 α + \frac{4 π}{3})

The Elegant Cancellation

Using the trigonometric identity

sin (A + B) + sin (A - B) = 2 sin A cos B

, the terms collapse beautifully. The sum simplifies to:

sin (2 α) + 2 sin (2 α) cos (\frac{2 π}{3})

Since

cos (\frac{2 π}{3}) = - \frac{1}{2}

, the expression becomes

sin (2 α) - sin (2 α) = 0

The normals are concurrent. This is a beautiful result and a testament to the harmony of geometry. Keep this result in your toolkit; it is a powerful weapon for your JEE journey.

Common Traps & FAQs

Wait, why do we use the eccentric angle

ϕ

instead of the actual angle

θ

from the center of the ellipse?

Great question! The eccentric angle

ϕ

is defined specifically for the ellipse using the auxiliary circle. The angle

θ

from the center (polar angle) doesn't have the same clean relationship with the normal equation. The auxiliary circle is the 'master key' for ellipse problems.

Why does the normal equation change when we multiply by

sin ϕ cos ϕ

It doesn't change the line itself, just the form. The standard form

a x sec ϕ - b y csc ϕ = a^{2} - b^{2}

is hard to work with when summing. Converting to

a x sin ϕ - b y cos ϕ = \frac{a ^{2} - b ^{2}}{2} sin 2 ϕ

makes it a linear equation in

x

and

y

that is much easier to manipulate algebraically.

Is the concurrency point always the same?

For a fixed equilateral triangle, yes. But if you rotate the triangle (change

α

), the concurrency point will move. The property is that for any

α

, the normals are concurrent, but the point of concurrency depends on

α

Visualized Solution

The Auxiliary Circle & Triangle

Eccentric Angles of $B$ and $C$

Projecting onto the Ellipse

Eccentric Angles of $P, Q, R$

Equation of the Normal

Rearranging the Normal Equation

Condition for Concurrency of Normals

Substituting the Angles

Setting up the Sum

Applying Trigonometric Identities

Final Evaluation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Bridge of Eccentric Angles

The Algebra of Normals

The Elegant Cancellation

Similar Questions

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.

Find the co-ordinates of all the points $P$ on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , for which the area of the triangle $P O N$ is a maximum, where $O$ denotes the origin and $N$ , the foot of the perpendicular from $O$ to the tangent at $P$ .

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.

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 :

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(\frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(- \frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

$C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the triangle $P A B$ lies on $C_{1}$ .

Visualized Solution

The Auxiliary Circle & Triangle

Eccentric Angles of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Projecting onto the Ellipse

Eccentric Angles of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P,Q,R

Equation of the Normal

Rearranging the Normal Equation

Condition for Concurrency of Normals

Substituting the Angles

Setting up the Sum

Applying Trigonometric Identities

Final Evaluation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Bridge of Eccentric Angles

The Algebra of Normals

The Elegant Cancellation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A tangent to the ellipse x2+4y2=4 meets the ellipse x2+2y2=6 at P and Q. Prove that the tangents at P and Q of the ellipse x2+2y2=6 are at right angles.

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.

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

.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 :

Eccentric Angles of $B$ and $C$

Eccentric Angles of $P, Q, R$

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.

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 :