JEE Advanced 2004

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Animated Solution for Mathematics - Conic Sections: 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

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

That is a fantastic question! While the $y = mx + c$ form is perfectly valid, it introduces a quadratic equation in terms of the slope $m$, which can make the algebra quite messy. The parametric form, using the angle $\theta$, allows us to handle the coordinates of the intercepts directly and elegantly. It simplifies the elimination of the parameter, which is the heart of finding a locus.

Q: How do I know if I should use $h$ and $k$ or just $x$ and $y$?

Think of $(h, k)$ as a specific, arbitrary point on the path we are tracing. We use $(h, k)$ to distinguish this point from the variables $x$ and $y$ that appear in the equation of the tangent line itself. Once we find the relationship between $h$ and $k$, we replace them with $x$ and $y$ to define the final path of the locus. It is a standard convention that keeps your work clean and logical.

Q: Is there a shortcut for the locus of the midpoint of the tangent intercept?

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the locus of the midpoint of the tangent intercept is indeed $\frac{a^2}{4x^2} + \frac{b^2}{4y^2} = 1$. While knowing this can be a great time-saver in competitive exams, I strongly encourage you to derive it as we did. Understanding the 'why' behind the formula ensures you can solve even more complex variations of this problem!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the tangent as a line that is constantly shifting its position as it touches the ellipse. The key to this problem is to realize that the midpoint of the tangent segment between the axes is a point

(h, k)

whose coordinates are functions of the angle

θ

that defines the point of tangency on the ellipse.

Hint 2

Hint 2 (The Arsenal): Recall the parametric equation of a tangent to the ellipse

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

, which is

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

. This equation is your most powerful tool for finding the intercepts on the coordinate axes.

Hint 3

Hint 3 (The Execution): Once you have the intercepts

A (\frac{a}{c o s θ}, 0)

and

B (0, \frac{b}{s i n θ})

, use the midpoint formula to express

h

and

k

in terms of

θ

. Finally, use the trigonometric identity

sin^{2} θ + cos^{2} θ = 1

to eliminate the parameter

θ

and find the locus.

Visualized Solution

Visualizing the Ellipse $x^{2} + 2 y^{2} = 2$

Given Ellipse: $x^{2} + 2 y^{2} = 2$
Objective: Find the locus of the midpoint of the tangent segment between the coordinate axes.

Standard Form of the Ellipse

Divide by $2$ to get standard form: $\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1$
Identify semi-axes: $a^{2} = 2 \Rightarrow a = 2$ and $b^{2} = 1 \Rightarrow b = 1$

Equation of the Tangent

Parametric form of tangent: $\frac{x c o s θ}{a} + \frac{y s i n θ}{b} = 1$
Substitute $a = 2$ and $b = 1$ : $\frac{x c o s θ}{2} + y sin θ = 1$

Finding the x-intercept $A$

For x-intercept (Point $A$ ), set $y = 0$ :
$x = \frac{2}{c o s θ} = 2 sec θ$
Point $A (2 sec θ, 0)$

Finding the y-intercept $B$

For y-intercept (Point $B$ ), set $x = 0$ :
$y = \frac{1}{s i n θ} = csc θ$
Point $B (0, csc θ)$

Defining the Midpoint $M (h, k)$

Let the midpoint of segment $A B$ be $M (h, k)$
As the tangent moves, point $M$ traces the required locus.

Applying the Midpoint Formula

By midpoint formula for $x$ -coordinate: $h = \frac{2 s e c θ + 0}{2} = \frac{s e c θ}{2}$
For $y$ -coordinate: $k = \frac{0 + c s c θ}{2} = \frac{c s c θ}{2}$

Rearranging for $θ$

To eliminate $θ$ , isolate the trigonometric terms:
$cos θ = \frac{1}{2 h}$
$sin θ = \frac{1}{2 k}$

Eliminating the Parameter $θ$

Use the fundamental identity: $sin^{2} θ + cos^{2} θ = 1$
Substitute the values: $(\frac{1}{2 k})^{2} + (\frac{1}{2 h})^{2} = 1$

The Final Locus Equation

Simplify the equation: $\frac{1}{4 k ^{2}} + \frac{1}{2 h ^{2}} = 1$
Replace $(h, k)$ with $(x, y)$ to get the final locus:
$\frac{1}{2 x ^{2}} + \frac{1}{4 y ^{2}} = 1$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given ellipse is defined by the equation

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

. To reveal its standard form, we divide the entire equation by

2

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

From this, we identify the semi-axes as

a^{2} = 2

(so

a = 2

) and

b^{2} = 1

(so

b = 1

The Tangent's Dance

A tangent line to the ellipse at a point defined by parameter

θ

is given by the equation:

\frac{x cos θ}{a} + \frac{y sin θ}{b} = 1

Substituting our specific values for

a

and

b

, the equation of the tangent becomes:

\frac{x cos θ}{2} + y sin θ = 1

This tangent intersects the coordinate axes at points

A

and

B

. Setting

y = 0

yields the

x

-intercept

A (2 sec θ, 0)

, and setting

x = 0

yields the

y

-intercept

B (0, csc θ)

The Midpoint's Journey

Let

M (h, k)

be the midpoint of the segment

A B

. Applying the midpoint formula, we obtain:

h = \frac{2 sec θ + 0}{2} = \frac{sec θ}{2}

k = \frac{0 + csc θ}{2} = \frac{csc θ}{2}

To determine the locus, we must eliminate the parameter

θ

. We rearrange the expressions to isolate the trigonometric functions:

cos θ = \frac{1}{2 h} and sin θ = \frac{1}{2 k}

The Final Synthesis

We utilize the fundamental trigonometric identity

sin^{2} θ + cos^{2} θ = 1

. Substituting our expressions for

sin θ

and

cos θ

into this identity, we get:

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

Simplifying the squares, we arrive at:

\frac{1}{4 k ^{2}} + \frac{1}{2 h ^{2}} = 1

Replacing

(h, k)

with the general coordinates

(x, y)

, we reach the final locus of the midpoint:

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

Common Traps & FAQs

Wait, why do we use the parametric form of the tangent instead of

y = m x + c

That is a fantastic question! While the

y = m x + c

form is perfectly valid, it introduces a quadratic equation in terms of the slope

m

, which can make the algebra quite messy. The parametric form, using the angle

θ

, allows us to handle the coordinates of the intercepts directly and elegantly. It simplifies the elimination of the parameter, which is the heart of finding a locus.

How do I know if I should use

h

and

k

or just

x

and

y

Think of

(h, k)

as a specific, arbitrary point on the path we are tracing. We use

(h, k)

to distinguish this point from the variables

x

and

y

that appear in the equation of the tangent line itself. Once we find the relationship between

h

and

k

, we replace them with

x

and

y

to define the final path of the locus. It is a standard convention that keeps your work clean and logical.

Is there a shortcut for the locus of the midpoint of the tangent intercept?

For an ellipse

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

, the locus of the midpoint of the tangent intercept is indeed

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

. While knowing this can be a great time-saver in competitive exams, I strongly encourage you to derive it as we did. Understanding the 'why' behind the formula ensures you can solve even more complex variations of this problem!

Animated Solution for Mathematics - Conic Sections: 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

Select Answer:

Visualized Solution

Visualizing the Ellipse $x^{2} + 2 y^{2} = 2$

Standard Form of the Ellipse

Equation of the Tangent

Finding the x-intercept $A$

Finding the y-intercept $B$

Defining the Midpoint $M (h, k)$

Applying the Midpoint Formula

Rearranging for $θ$

Eliminating the Parameter $θ$

The Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

The Midpoint's Journey

The Final Synthesis

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 :

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

Consider the family of circles $x^{2} + y^{2} = r^{2}, 2 < r < 5$ . If in the first quadrant, the common tangent to a circle of this family and the ellipse $4 x^{2} + 25 y^{2} = 100$ meets the co-ordinate axes at $A$ and $B$ , then find the equation of the locus of the mid-point of $A B$ .

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the mid point of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points

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?

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2} + 3 y^{2} = 6$ on any tangent to it 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 :

Select Answer:

Visualized Solution

Visualizing the Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+2y2=2

Standard Form of the Ellipse

Equation of the Tangent

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

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

Defining the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(h,k)

Applying the Midpoint Formula

Rearranging for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

The Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

The Midpoint's Journey

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn to the hyperbola 4y2=x2+1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

.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; } From the origin chords are drawn to the circle (x−1)2+y2=1. The equation of the locus of the mid-points of these chords is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :

Visualizing the Ellipse $x^{2} + 2 y^{2} = 2$

Finding the x-intercept $A$

Finding the y-intercept $B$

Defining the Midpoint $M (h, k)$

Rearranging for $θ$

Eliminating the Parameter $θ$

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

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

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?

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2} + 3 y^{2} = 6$ on any tangent to it 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 :