JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

Q: Why can't we just use the Cartesian equation of the ellipse directly? It seems like more work to use $\theta$.

Great question! If you use $y = \sqrt{2 - x^2/2}$, you will end up with square roots and messy algebra that will make your head spin. Parametric coordinates allow us to use the identity $\cos^2 \theta + \sin^2 \theta = 1$, which is the most elegant way to eliminate the variable and find the locus.

Q: Does the fixed point $P(4, 3)$ affect the eccentricity of the resulting ellipse?

Actually, no! The fixed point $P$ acts as a translation factor. It shifts the center of the new ellipse, but it does not change the ratio of the semi-major and semi-minor axes. The eccentricity is a property of the shape's 'flatness,' not its position in the coordinate plane.

Q: How do I know if the resulting locus is definitely an ellipse?

Look at the final equation: $(h - 2)^2 + 2(k - 3/2)^2 = 1$. Because the coefficients of the squared terms are positive and different, it fits the standard form $\frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1$. Since $A^2 \neq B^2$, it is guaranteed to be an ellipse.

Select Answer:

Visualized Solution

Visualizing the Setup

Given Ellipse: $x^{2} + 2 y^{2} = 4$
Fixed Point: $P (4, 3)$
Objective: Find the eccentricity of the locus of the midpoint of $P Q$ , where $Q$ is any point on the ellipse.

Standard Form of the Ellipse

Divide the equation $x^{2} + 2 y^{2} = 4$ by $4$ :
$\frac{x ^{2}}{4} + \frac{2 y ^{2}}{4} = \frac{4}{4}$
Standard Form: $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$
Here, $a^{2} = 4$ and $b^{2} = 2$ .

Parametric Coordinates of Point $Q$

Any point $Q$ on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is $(a cos θ, b sin θ)$ .
For our ellipse: $a = 2, b = 2$
Coordinates of $Q$ : $(2 cos θ, 2 sin θ)$

Defining the Midpoint $M (h, k)$

Let $M (h, k)$ be the midpoint of $P (4, 3)$ and $Q (2 cos θ, 2 sin θ)$ .
Draw the line segment $P Q$ .
The midpoint $M$ will trace a path as $Q$ moves.

Applying the Midpoint Formula

Using Midpoint Formula: $h = \frac{x _{1} + x _{2}}{2}, k = \frac{y _{1} + y _{2}}{2}$
$h = \frac{4 + 2 c o s θ}{2}$
$k = \frac{3 + 2 s i n θ}{2}$

Isolating $cos θ$

From $h = \frac{4 + 2 c o s θ}{2}$ :
$2 h = 4 + 2 cos θ$
$2 h - 4 = 2 cos θ$
$cos θ = h - 2$

Isolating $sin θ$

From $k = \frac{3 + 2 s i n θ}{2}$ :
$2 k = 3 + 2 sin θ$
$2 k - 3 = 2 sin θ$
$sin θ = \frac{2 k - 3}{2}$

Eliminating the Parameter $θ$

Use the fundamental trigonometric identity: $cos^{2} θ + sin^{2} θ = 1$
Substitute the isolated expressions:
$(h - 2)^{2} + (\frac{2 k - 3}{2})^{2} = 1$

Simplifying the Locus Equation

Expand the squared term for $k$ :
$(h - 2)^{2} + \frac{( 2 k - 3 ) ^{2}}{2} = 1$
Factor out $2$ from the $k$ term inside the square:
$(h - 2)^{2} + \frac{4 ( k - \frac{3}{2} ) ^{2}}{2} = 1$
$(h - 2)^{2} + 2 (k - \frac{3}{2})^{2} = 1$

Standard Form of the Locus

Replace $(h, k)$ with $(x, y)$ to get the general locus:
$\frac{( x - 2 ) ^{2}}{1} + \frac{( y - \frac{3}{2} ) ^{2}}{\frac{1}{2}} = 1$
This is an ellipse of the form $\frac{( x - x _{0} ) ^{2}}{A ^{2}} + \frac{( y - y _{0} ) ^{2}}{B ^{2}} = 1$
Where $A^{2} = 1$ and $B^{2} = \frac{1}{2}$

Calculating Eccentricity

Eccentricity formula: $e = 1 - \frac{B ^{2}}{A ^{2}}$
Substitute $A^{2} = 1$ and $B^{2} = \frac{1}{2}$ :
$e = 1 - \frac{\frac{1}{2}}{1}$
$e = 1 - \frac{1}{2} = \frac{1}{2}$
$e = \frac{1}{2}$

Conclusion

Final Answer: The eccentricity of the locus is $e = \frac{1}{2}$ .
Key Takeaway: The locus of the midpoint of a segment joining a fixed point and a point on an ellipse is another ellipse with the same eccentricity as the original one!
Notice the original ellipse had $e = 1 - 2/4 = \frac{1}{2}$ .

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

Solution Diagram

Analyzing the Setup

The given ellipse is defined by the equation

x^{2} + 2 y^{2} = 4

. To understand its geometry, we divide the entire equation by

4

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

From this standard form, we identify the semi-major axis

a = 2

and the semi-minor axis

b = 2

. This establishes our foundation for the moving point

Q

on the perimeter.

The Parametric Dance

To track the motion of point

Q

effectively, we utilize the parametric representation

Q (a cos θ, b sin θ)

. Substituting our known values, the coordinates of

Q

are:

Q (2 cos θ, 2 sin θ)

As the parameter

θ

varies from

0

2 π

, point

Q

traces the entire boundary of the ellipse. This parameterization is essential for eliminating the constraint during the derivation of the locus.

The Midpoint Bridge

Let

M (h, k)

be the midpoint of the segment connecting the fixed point

P (4, 3)

and the moving point

Q (2 cos θ, 2 sin θ)

. Using the midpoint formula, we define the coordinates of

M

as:

h = \frac{4 + 2 cos θ}{2} = 2 + cos θ

k = \frac{3 + 2 sin θ}{2}

These equations link the locus coordinates

(h, k)

to the parameter

θ

. Our goal is to eliminate

θ

to find the path traced by

M

Algebraic Alchemy

We isolate the trigonometric functions from the midpoint equations:

cos θ = h - 2

sin θ = \frac{2 k - 3}{2}

Applying the fundamental trigonometric identity

cos^{2} θ + sin^{2} θ = 1

, we substitute our expressions:

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

Expanding the second term, we obtain:

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

Final Calculation

To express the locus in standard form, we factor out the coefficient of

k

(h - 2)^{2} + \frac{4 ( k - 3/2 ) ^{2}}{2} = 1

(h - 2)^{2} + 2 (k - 3/2)^{2} = 1

Replacing

(h, k)

with

(x, y)

, the final equation of the locus is:

\frac{( x - 2 ) ^{2}}{1} + \frac{( y - 3/2 ) ^{2}}{1/2} = 1

This result confirms that the locus is an ellipse centered at

(2, 3/2)

. The eccentricity

e

of this new ellipse is calculated as:

e = 1 - \frac{1/2}{1} = \frac{1}{2}

This demonstrates that the midpoint transformation preserves the eccentricity of the original conic section.

Common Traps & FAQs

Why can't we just use the Cartesian equation of the ellipse directly? It seems like more work to use

θ

Great question! If you use

y = 2 - x^{2} /2

, you will end up with square roots and messy algebra that will make your head spin. Parametric coordinates allow us to use the identity

cos^{2} θ + sin^{2} θ = 1

, which is the most elegant way to eliminate the variable and find the locus.

Does the fixed point

P (4, 3)

affect the eccentricity of the resulting ellipse?

Actually, no! The fixed point

P

acts as a translation factor. It shifts the center of the new ellipse, but it does not change the ratio of the semi-major and semi-minor axes. The eccentricity is a property of the shape's 'flatness,' not its position in the coordinate plane.

How do I know if the resulting locus is definitely an ellipse?

Look at the final equation:

(h - 2)^{2} + 2 (k - 3/2)^{2} = 1

. Because the coefficients of the squared terms are positive and different, it fits the standard form

\frac{( x - h ) ^{2}}{A ^{2}} + \frac{( y - k ) ^{2}}{B ^{2}} = 1

. Since

A^{2} \neq = B^{2}

, it is guaranteed to be an ellipse.

Animated Solution for Mathematics - Conic Sections: The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

Select Answer:

Visualized Solution

Visualizing the Setup

Standard Form of the Ellipse

Parametric Coordinates of Point $Q$

Defining the Midpoint $M (h, k)$

Applying the Midpoint Formula

Isolating $cos θ$

Isolating $sin θ$

Eliminating the Parameter $θ$

Simplifying the Locus Equation

Standard Form of the Locus

Calculating Eccentricity

Conclusion

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

Analyzing the Setup

The Parametric Dance

The Midpoint Bridge

Algebraic Alchemy

Final Calculation

Similar Questions

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 :

The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units, then its eccentricity is :

Select Answer:

Visualized Solution

Visualizing the Setup

Standard Form of the Ellipse

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

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

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

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

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

Simplifying the Locus Equation

Standard Form of the Locus

Calculating Eccentricity

Conclusion

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

Analyzing the Setup

The Parametric Dance

The Midpoint Bridge

Algebraic Alchemy

Final Calculation

Similar Questions

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the point of intersection of the lines (3​)kx+ky−43​=0 and 3​x−y−4(3​)k=0 is a conic, whose eccentricity is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the point of intersection of the lines, 2​x−y+42​k=0 and 2​kx+ky−42​=0 (k is any non-zero real parameter), is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1:3. Then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrices is x=4, then the equation of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the eccentricity of an ellipse a2x2​+b2y2​=1,a>b, be 41​. If this ellipse passes through the point (−452​​,3), then a2+b2 is equal to :

Parametric Coordinates of Point $Q$

Defining the Midpoint $M (h, k)$

Isolating $cos θ$

Isolating $sin θ$

Eliminating the Parameter $θ$

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 :

The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :