JEE Advanced 2002

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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.

Q: Why do we use the $T=0$ method for the tangent? Can't we just use calculus?

You absolutely can use calculus! Differentiating $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives you the slope $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$. However, the $T=0$ method is a powerful, elegant shortcut in coordinate geometry that saves you precious time during the JEE Advanced exam. It is a standard tool that every top-tier student should master to keep their workflow clean and error-free.

Progressive Hints

Hint 1

Visualize the geometry: You are looking for the intersection of two lines. The directrix is defined by the vertical line

x = \frac{a}{e}

. Your goal is to prove that the intersection point of the perpendicular from the focus and the line joining the center to the point of contact always has an x-coordinate of

\frac{a}{e}

Hint 2

Recall the equation of the tangent at

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

using the

T = 0

method, which is

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

. From this, determine the slope

m_{1}

. Then, use the condition for perpendicular lines,

m_{1} \cdot m_{2} = - 1

, to find the slope

m_{2}

of the line passing through the focus

S (a e, 0)

Hint 3

Set up the equation for the perpendicular line passing through

S (a e, 0)

and the line

O P

passing through the origin

(0, 0)

and

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

. Equate the two expressions for

y

to find the intersection point. Finally, use the fundamental ellipse identity

b^{2} = a^{2} (1 - e^{2})

to simplify the expression and reveal the x-coordinate as

\frac{a}{e}

Visualized Solution

Setting the Stage: The Ellipse and its Components

Standard Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Focus: $S (a e, 0)$
Point of contact: $P (x_{1}, y_{1})$
Directrix: $x = \frac{a}{e}$

Equation of the Tangent at $P (x_{1}, y_{1})$

Equation of tangent at $P (x_{1}, y_{1})$ : $\frac{x x _{1}}{a ^{2}} + \frac{y y _{1}}{b ^{2}} = 1$
Slope of tangent ( $m_{1}$ ): $- \frac{b ^{2} x _{1}}{a ^{2} y _{1}}$

Finding the Perpendicular from Focus $S$

Slope of perpendicular ( $m_{2}$ ): $\frac{a ^{2} y _{1}}{b ^{2} x _{1}}$
Equation of perpendicular from $S (a e, 0)$ : $y - 0 = \frac{a ^{2} y _{1}}{b ^{2} x _{1}} (x - a e)$

Equation of the Line Joining Center to $P$

Line $O P$ joins $(0, 0)$ and $(x_{1}, y_{1})$
Equation of line $O P$ : $y = \frac{y _{1}}{x _{1}} x$

Finding the Intersection Point

Equating $y$ from both equations:
$\frac{y _{1}}{x _{1}} x = \frac{a ^{2} y _{1}}{b ^{2} x _{1}} (x - a e)$

Algebraic Simplification

Cancel $\frac{y _{1}}{x _{1}}$ from both sides: $x = \frac{a ^{2}}{b ^{2}} (x - a e)$
Multiply by $b^{2}$ : $b^{2} x = a^{2} x - a^{3} e$
Rearrange: $(a^{2} - b^{2}) x = a^{3} e$

Using Ellipse Properties

Property: $b^{2} = a^{2} (1 - e^{2}) \Rightarrow a^{2} - b^{2} = a^{2} e^{2}$
Substitute: $a^{2} e^{2} x = a^{3} e$

Conclusion: Meeting on the Directrix

Solve for $x$ : $x = \frac{a ^{3} e}{a ^{2} e ^{2}} = \frac{a}{e}$
This is the equation of the directrix.
Hence Proved: The lines meet on the directrix.

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We consider the standard ellipse defined by the equation:

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

The focus

S

is located at

(a e, 0)

, and the directrix is the vertical line

x = \frac{a}{e}

. We select an arbitrary point

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

on the ellipse to serve as our anchor.

The Tangent and the Perpendicular Line

Using the

T = 0

method, the equation of the tangent at

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

is:

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

Rearranging this into slope-intercept form, we find the slope

m_{1}

of the tangent:

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

Since the line drawn from the focus

S

is perpendicular to this tangent, its slope

m_{2}

must satisfy

m_{1} \cdot m_{2} = - 1

. Thus, we calculate:

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

Using the point-slope form with the focus

S (a e, 0)

, the equation of this perpendicular line is:

y = \frac{a ^{2} y _{1}}{b ^{2} x _{1}} (x - a e)

The Line from the Center

Next, we consider the line

O P

connecting the center

(0, 0)

to the point

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

. The equation for this line is:

y = \frac{y _{1}}{x _{1}} x

The Intersection Point

To find the intersection of the perpendicular line and the line

O P

, we set their

y

-values equal:

\frac{y _{1}}{x _{1}} x = \frac{a ^{2} y _{1}}{b ^{2} x _{1}} (x - a e)

We can cancel the common term

\frac{y _{1}}{x _{1}}

from both sides, yielding:

x = \frac{a ^{2}}{b ^{2}} (x - a e)

Final Calculation

Multiplying both sides by

b^{2}

gives

b^{2} x = a^{2} x - a^{3} e

, which rearranges to:

(a^{2} - b^{2}) x = a^{3} e

Recalling the fundamental identity of the ellipse,

a^{2} - b^{2} = a^{2} e^{2}

, we substitute this into the equation:

a^{2} e^{2} x = a^{3} e

Dividing both sides by

a^{2} e^{2}

, we arrive at the final result:

$x = \frac{a}{e}$

This confirms that the intersection point lies on the directrix, proving the theorem.

Common Traps & FAQs

Why do we use the

T = 0

method for the tangent? Can't we just use calculus?

You absolutely can use calculus! Differentiating

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

gives you the slope

\frac{d y}{d x} = - \frac{b ^{2} x}{a ^{2} y}

. However, the

T = 0

method is a powerful, elegant shortcut in coordinate geometry that saves you precious time during the JEE Advanced exam. It is a standard tool that every top-tier student should master to keep their workflow clean and error-free.

Why does the term

\frac{y _{1}}{x _{1}}

cancel out during the intersection step? Is that a coincidence?

It is not a coincidence at all—it is the beauty of the ellipse! The fact that

\frac{y _{1}}{x _{1}}

cancels out means that the intersection point's x-coordinate is independent of the specific position of point

P

on the ellipse. This proves that the property holds true for *any* tangent you choose. It is a universal geometric truth, not a specific case.

Can I use parametric coordinates

(a cos θ, b sin θ)

instead of

(x_{1}, y_{1})

Yes, you can! Using parametric coordinates often simplifies the algebra significantly because it removes the need to carry around the

x_{1}

and

y_{1}

variables. If you feel comfortable with trigonometry, go for it. The result will be identical, and you might find the path to the solution even faster.

Animated Solution for Mathematics - Conic Sections: 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.

Visualized Solution

Setting the Stage: The Ellipse and its Components

Equation of the Tangent at $P (x_{1}, y_{1})$

Finding the Perpendicular from Focus $S$

Equation of the Line Joining Center to $P$

Finding the Intersection Point

Algebraic Simplification

Using Ellipse Properties

Conclusion: Meeting on the Directrix

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent and the Perpendicular Line

The Line from the Center

The Intersection Point

Final Calculation

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.

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1$ meet at the point $R (2, 22 - 2)$ . If $S$ is the focus of the ellipse on its negative major axis, then $S P^{2} + S Q^{2}$ is equal to

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ from any of its foci?

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

Animated Solution for Mathematics - Conic Sections: 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.

Visualized Solution

Setting the Stage: The Ellipse and its Components

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x1​,y1​)

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

Equation of the Line Joining Center to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding the Intersection Point

Algebraic Simplification

Using Ellipse Properties

Conclusion: Meeting on the Directrix

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent and the Perpendicular Line

The Line from the Center

The Intersection Point

Final Calculation

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.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, 4x2​+2y2​=1 from any of its foci?

.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

Equation of the Tangent at $P (x_{1}, y_{1})$

Finding the Perpendicular from Focus $S$

Equation of the Line Joining Center to $P$

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.

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ from any of its foci?

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

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