JEE Advanced 2005

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Tangents are drawn from any point on the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ to the circle $x^{2} + y^{2} = 9$ . Find the locus of mid-point of the chord of contact.

Q: Why do we equate the ratios of the coefficients? Is that always valid?

Yes, it is! When you have two linear equations, say $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, and you know they represent the exact same line, it means one equation is just a scalar multiple of the other. Therefore, the ratios of their coefficients must be equal: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. This is a fundamental tool in coordinate geometry for eliminating parameters and finding loci.

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the geometry. You have a hyperbola and a circle. The chord of contact is defined by the tangents from a point

P

on the hyperbola to the circle. But remember, this chord also has a midpoint

M (h, k)

. The key is to link the definition of the chord of contact from

P

with the definition of a chord with a known midpoint

M

Hint 2

Hint 2: The Arsenal. You need two standard equations for the same line. First, the equation of the chord of contact from an external point

(x_{1}, y_{1})

to the circle

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

T = 0

, which is

x x_{1} + y y_{1} = r^{2}

. Second, the equation of a chord with a given midpoint

(h, k)

T = S_{1}

, which is

x h + y k = h^{2} + k^{2}

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

Hint 3

Hint 3: The Execution. Set up the proportionality:

\frac{3 s e c θ}{h} = \frac{2 t a n θ}{k} = \frac{9}{h ^{2} + k ^{2}}

. From this, isolate

sec θ

and

tan θ

in terms of

h

and

k

. Finally, use the fundamental identity

sec^{2} θ - tan^{2} θ = 1

to eliminate the parameter

θ

and find the locus.

Visualized Solution

Visualizing the Setup

Hyperbola: $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$
Circle: $x^{2} + y^{2} = 9$
Observe that the vertices of the hyperbola $(\pm 3, 0)$ lie exactly on the circle.

Parametric Point $P$

Let $P (x_{1}, y_{1})$ be any point on the hyperbola.
Using parametric coordinates, we can write $P$ as $(3 sec θ, 2 tan θ)$ .
Here, $θ$ is the parameter that varies as $P$ moves along the curve.

Tangents and Chord of Contact

From point $P$ , we draw two tangents to the circle.
The line joining the points of tangency is the Chord of Contact.
The equation of the chord of contact from $(x_{1}, y_{1})$ to $x^{2} + y^{2} = r^{2}$ is given by $T = 0$ .

Equation of Chord of Contact

Substitute $P (3 sec θ, 2 tan θ)$ into $T = 0$ for the circle $x^{2} + y^{2} = 9$ .
The expression for $T$ is $x x_{1} + y y_{1} - r^{2}$ .
Equation: $(3 sec θ) x + (2 tan θ) y = 9$ .

Midpoint Form $T = S_{1}$

We need the locus of the midpoint of this chord. Let the midpoint be $M (h, k)$ .
The equation of a chord with a given midpoint $(h, k)$ is $T = S_{1}$ .

Equation of Chord with Midpoint

Apply $T = S_{1}$ for the circle $x^{2} + y^{2} = 9$ using midpoint $(h, k)$ .
$T$ becomes $h x + k y - 9$ .
$S_{1}$ becomes $h^{2} + k^{2} - 9$ .
Equating them: $h x + k y = h^{2} + k^{2}$ .

Comparing the Equations

We have two equations for the same chord:
1. $(3 sec θ) x + (2 tan θ) y = 9$
2. $h x + k y = h^{2} + k^{2}$
Since they represent the same line, their coefficients must be proportional.
$\frac{3 s e c θ}{h} = \frac{2 t a n θ}{k} = \frac{9}{h ^{2} + k ^{2}}$

Isolating $sec θ$ and $tan θ$

From the ratios, we can express $sec θ$ and $tan θ$ in terms of $h$ and $k$ .
$sec θ = \frac{9 h}{3 ( h ^{2} + k ^{2} )} = \frac{3 h}{h ^{2} + k ^{2}}$
$tan θ = \frac{9 k}{2 ( h ^{2} + k ^{2} )}$

Eliminating the Parameter

To eliminate $θ$ , we use the fundamental trigonometric identity:
$sec^{2} θ - tan^{2} θ = 1$
This will link $h$ and $k$ directly, giving us the locus equation.

Substituting into Identity

Substitute the expressions for $sec θ$ and $tan θ$ into the identity.
$(\frac{3 h}{h ^{2} + k ^{2}})^{2} - (\frac{9 k}{2 ( h ^{2} + k ^{2} )})^{2} = 1$

Expanding the Squares

Expand the squared terms carefully:
$\frac{9 h ^{2}}{( h ^{2} + k ^{2} ) ^{2}} - \frac{81 k ^{2}}{4 ( h ^{2} + k ^{2} ) ^{2}} = 1$
Notice that the denominator $(h^{2} + k^{2})^{2}$ is common to both terms on the left side.

Rearranging the Equation

Divide the entire equation by $81$ :
$\frac{h ^{2}}{9 ( h ^{2} + k ^{2} ) ^{2}} - \frac{k ^{2}}{4 ( h ^{2} + k ^{2} ) ^{2}} = \frac{1}{81}$
Multiply both sides by $(h^{2} + k^{2})^{2}$ :
$\frac{h ^{2}}{9} - \frac{k ^{2}}{4} = \frac{( h ^{2} + k ^{2} ) ^{2}}{81}$

The Final Locus

Rewrite the right side: $\frac{( h ^{2} + k ^{2} ) ^{2}}{81} = (\frac{h ^{2} + k ^{2}}{9})^{2}$
To get the final locus, replace the dummy coordinates $(h, k)$ with $(x, y)$ :
$\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = (\frac{x ^{2} + y ^{2}}{9})^{2}$
This represents the path traced by the midpoint $M$ .

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The problem involves a hyperbola defined by the equation:

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

The vertices of this hyperbola lie on the circle

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

. We consider a point

P

on the hyperbola and draw tangents from

P

to the circle. We aim to find the locus of the midpoint

M (h, k)

of the chord of contact.

The Power of Parametric Coordinates

To simplify the geometry, we represent any point

P

on the hyperbola using parametric coordinates:

P = (3 sec θ, 2 tan θ)

As the parameter

θ

varies,

P

traces the entire hyperbola. This parameter will serve as the bridge between the position of

P

and the geometry of the chord.

The Dual Nature of the Chord

The equation of the chord of contact from an external point

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

to the circle

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

is given by

T = 0

, which is

x x_{1} + y y_{1} = 9

. Substituting our parametric coordinates for

P

, we obtain:

(3 sec θ) x + (2 tan θ) y = 9

Alternatively, if

M (h, k)

is the midpoint of a chord, the equation of that chord is given by

T = S_{1}

. For our circle, this results in:

h x + k y = h^{2} + k^{2}

The Synthesis

Since both equations represent the same line, their coefficients must be proportional. This yields the following relationship:

\frac{3 sec θ}{h} = \frac{2 tan θ}{k} = \frac{9}{h ^{2} + k ^{2}}

From this ratio, we can isolate

sec θ

and

tan θ

in terms of the coordinates of the midpoint

(h, k)

sec θ = \frac{9 h}{3 ( h ^{2} + k ^{2} )} = \frac{3 h}{h ^{2} + k ^{2}}

tan θ = \frac{9 k}{2 ( h ^{2} + k ^{2} )}

The Final Elimination

To find the locus, we eliminate

θ

using the fundamental trigonometric identity

sec^{2} θ - tan^{2} θ = 1

. Substituting our expressions into this identity gives:

(\frac{3 h}{h ^{2} + k ^{2}})^{2} - (\frac{9 k}{2 ( h ^{2} + k ^{2} )})^{2} = 1

Expanding the terms, we have:

\frac{9 h ^{2}}{( h ^{2} + k ^{2} ) ^{2}} - \frac{81 k ^{2}}{4 ( h ^{2} + k ^{2} ) ^{2}} = 1

Multiplying the entire equation by

4 (h^{2} + k^{2})^{2}

to clear the denominators, we obtain:

36 h^{2} - 81 k^{2} = 4 (h^{2} + k^{2})^{2}

Replacing

(h, k)

with the general coordinates

(x, y)

, we arrive at the final locus:

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = (\frac{x ^{2} + y ^{2}}{9})^{2}

This equation represents the locus of the midpoint of the chord of contact.

Common Traps & FAQs

Wait, why can we use

T = S_{1}

for the chord of contact? Isn't that only for chords with a given midpoint?

That is a fantastic question! You are absolutely right that

T = S_{1}

is the standard form for a chord with a given midpoint

(h, k)

. However, the chord of contact *is* a chord, and it *does* have a midpoint. By calling that midpoint

(h, k)

, we are simply using the properties of that specific chord. Since the chord of contact is a unique line, it must satisfy the

T = S_{1}

condition for its own midpoint. It is a brilliant way to bridge the two different ways of defining the same line!

Why do we equate the ratios of the coefficients? Is that always valid?

Yes, it is! When you have two linear equations, say

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, and you know they represent the exact same line, it means one equation is just a scalar multiple of the other. Therefore, the ratios of their coefficients must be equal:

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

. This is a fundamental tool in coordinate geometry for eliminating parameters and finding loci.

Is the final locus supposed to be a standard conic section? This looks like a higher-degree curve.

You have a sharp eye! In many JEE problems, the locus turns out to be a standard conic like a parabola or ellipse. But here, because the relationship involves the square of the distance from the origin, the resulting locus is a higher-degree curve. Do not be alarmed by the complexity of the final equation. In advanced geometry, the 'path' is often more complex than the simple shapes we start with. Trust your algebraic derivation!

Animated Solution for Mathematics - Conic Sections: Tangents are drawn from any point on the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ to the circle $x^{2} + y^{2} = 9$ . Find the locus of mid-point of the chord of contact.

Visualized Solution

Visualizing the Setup

Parametric Point $P$

Tangents and Chord of Contact

Equation of Chord of Contact

Midpoint Form $T = S_{1}$

Equation of Chord with Midpoint

Comparing the Equations

Isolating $sec θ$ and $tan θ$

Eliminating the Parameter

Substituting into Identity

Expanding the Squares

Rearranging the Equation

The Final Locus

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

Analyzing the Setup

The Power of Parametric Coordinates

The Dual Nature of the Chord

The Synthesis

The Final Elimination

Similar Questions

Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Equation of the circle with $A B$ as its diameter is

From a point $A$ common tangents are drawn to the circle $x^{2} + y^{2} = a^{2} /2$ and parabola $y^{2} = 4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

Let $S_{1} : x^{2} + y^{2} = 9$ and $S_{2} : (x - 2)^{2} + y^{2} = 1$ . Then the locus of center of a variable circle $S$ which touches $S_{1}$ internally and $S_{2}$ externally always passes through the points:

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

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 :

A line is a common tangent to the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ . If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2 (a + c)$ is equal to

The locus of mid-points of the line segments joining $(- 3, - 5)$ and the points on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ is :

If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ and a point P moves on the line $2 x - 3 y + 4 = 0$ then the centroid of $△ P A B$ lies on the line :

Visualized Solution

Visualizing the Setup

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

Tangents and Chord of Contact

Equation of Chord of Contact

Midpoint Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​

Equation of Chord with Midpoint

Comparing the Equations

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } secθ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ

Eliminating the Parameter

Substituting into Identity

Expanding the Squares

Rearranging the Equation

The Final Locus

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

Analyzing the Setup

The Power of Parametric Coordinates

The Dual Nature of the Chord

The Synthesis

The Final Elimination

Similar Questions

Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the circle with AB as its diameter is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=0, and its foci is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Show that the locus of a point that divides a chord of slope 2 of the parabola y2=4x internally in the ratio 1:2 is a parabola. Find the vertex of this parabola.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of mid-points of the line segments joining (−3,−5) and the points on the ellipse 4x2​+9y2​=1 is :

Parametric Point $P$

Midpoint Form $T = S_{1}$

Isolating $sec θ$ and $tan θ$

Equation of the circle with $A B$ as its diameter is

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

The locus of mid-points of the line segments joining $(- 3, - 5)$ and the points on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ is :