JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre $C$ of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius $r$ . Let $f_{1}, f_{2}$ be the focal distances of the point $C$ on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to

Q: Why do we need to find $p$ first? Can't we just work with the ellipse equation as is?

You cannot proceed without $p$ because the ellipse is not fully defined! The problem gives you the ellipse $3x^2 + py^2 = 4$, but $p$ is an unknown. Since the problem tells us the ellipse passes through the center of the circle, that point must satisfy the equation. Finding $p$ is the key that unlocks the entire problem. Without it, you do not have a specific ellipse to work with.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark: Visualize the geometry. You have a circle and an ellipse that share a specific point of intersection, the center of the circle

C (1, 2)

. This point is the bridge between the two curves. Your first task is to extract the coordinates of

C

and the radius

r

from the circle equation, then use the fact that

C

lies on the ellipse to find the unknown parameter

p

Hint 2

Hint 2: The Arsenal: Once you have the ellipse equation, you must determine its orientation. Compare the coefficients of

x^{2}

and

y^{2}

to decide if it is a horizontal or vertical ellipse. Remember, for a vertical ellipse, the focal distances of a point

(x, y)

are given by

b \pm ey

. This is a powerful property that saves you from calculating the coordinates of the foci directly.

Hint 3

Hint 3: The Execution: Start by calculating the eccentricity

e

using the relation

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

for a vertical ellipse. Once you have

e^{2}

, use the product of the focal distances formula:

f_{1} f_{2} = (b + ey) (b - ey) = b^{2} - e^{2} y^{2}

. Substitute the values of

b^{2}

e^{2}

, and the

y

-coordinate of the center to find the product, then compute

6 f_{1} f_{2} - r

Visualized Solution

Center and Radius of the Circle

Circle equation: $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$
Compare with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Center $C = (- g, - f) = (1, 2)$
Radius $r = g^{2} + f^{2} - c = 1 + 4 + 11 = 4$

Finding the Unknown Parameter $p$

The ellipse $3 x^{2} + p y^{2} = 4$ passes through the center $C (1, 2)$ .
Substitute $x = 1$ and $y = 2$ into the ellipse equation.
$3 (1)^{2} + p (2)^{2} = 4$
$3 + 4 p = 4 ⟹ 4 p = 1 ⟹ p = \frac{1}{4}$

Standard Form of the Ellipse

Substitute $p = \frac{1}{4}$ back into the ellipse equation: $3 x^{2} + \frac{y ^{2}}{4} = 4$
Divide the entire equation by $4$ to make the right side $1$ .
$\frac{3 x ^{2}}{4} + \frac{y ^{2}}{16} = 1$
Rewrite as: $\frac{x ^{2}}{4/3} + \frac{y ^{2}}{16} = 1$

Identifying the Major Axis

Compare with standard form: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
We get $a^{2} = \frac{4}{3}$ and $b^{2} = 16$ .
Since $b^{2} > a^{2}$ , the major axis lies along the y-axis.
This is a vertical ellipse.

Finding the Eccentricity

For a vertical ellipse ( $b > a$ ), the eccentricity formula is $a^{2} = b^{2} (1 - e^{2})$ .
Rearranging for $e^{2}$ : $e^{2} = 1 - \frac{a ^{2}}{b ^{2}}$
Substitute the values: $e^{2} = 1 - \frac{4/3}{16} = 1 - \frac{4}{48}$
$e^{2} = 1 - \frac{1}{12} = \frac{11}{12}$

Formula for Focal Distances

Let $f_{1}$ and $f_{2}$ be the distances from point $C (1, 2)$ to the foci.
For a point $(x, y)$ on a vertical ellipse, the focal distances are $b + ey$ and $b - ey$ .
Therefore, $f_{1} = b + ey$ and $f_{2} = b - ey$ .
We need to find the product $f_{1} f_{2}$ .

Computing the Product of Focal Distances

Multiply the focal distances: $f_{1} f_{2} = (b + ey) (b - ey) = b^{2} - e^{2} y^{2}$
We know $b^{2} = 16$ , $e^{2} = \frac{11}{12}$ , and for point $C (1, 2)$ , $y = 2$ .
Substitute these values: $f_{1} f_{2} = 16 - (\frac{11}{12}) (2)^{2}$
$f_{1} f_{2} = 16 - (\frac{11}{12}) (4) = 16 - \frac{11}{3}$
$f_{1} f_{2} = \frac{48 - 11}{3} = \frac{37}{3}$

Evaluating the Final Expression

We need to find the value of $6 f_{1} f_{2} - r$ .
We found $f_{1} f_{2} = \frac{37}{3}$ and $r = 4$ .
Substitute the values: $6 (\frac{37}{3}) - 4$
Simplify: $2 (37) - 4 = 74 - 4$
Final Answer $= 70$

00:00 / 00:00

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

Solution Diagram

Analyzing the Circle

We begin by examining the circle defined by the equation:

x^{2} + y^{2} - 2 x - 4 y - 11 = 0

By completing the square or comparing it to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center

C

(1, 2)

. The radius

r

is calculated as:

r = 1^{2} + 2^{2} - (- 11) = 16 = 4

Determining the Ellipse Parameters

The ellipse is given by the equation

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

. Since the ellipse passes through the center of the circle

C (1, 2)

, these coordinates must satisfy the equation:

3 (1)^{2} + p (2)^{2} = 4 \Rightarrow 3 + 4 p = 4 \Rightarrow p = \frac{1}{4}

Substituting

p = 1/4

back into the equation, we obtain

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

. Dividing by

4

yields the standard form:

\frac{x ^{2}}{4/3} + \frac{y ^{2}}{16} = 1

Geometry of the Ellipse

Comparing this to the standard form

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

, we identify

a^{2} = 4/3

and

b^{2} = 16

. Since

b^{2} > a^{2}

, the ellipse is vertical.

We calculate the eccentricity

e

using the relation

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

e^{2} = 1 - \frac{a ^{2}}{b ^{2}} = 1 - \frac{4/3}{16} = 1 - \frac{1}{12} = \frac{11}{12}

Calculating Focal Distances

For a vertical ellipse, the focal distances

f_{1}

and

f_{2}

for any point

(x, y)

are given by

b \pm ey

. The product of these distances is:

f_{1} f_{2} = (b + ey) (b - ey) = b^{2} - e^{2} y^{2}

Substituting

b^{2} = 16

e^{2} = 11/12

, and the

y

-coordinate of the center

y = 2

f_{1} f_{2} = 16 - (\frac{11}{12}) (2)^{2} = 16 - (\frac{11}{12}) (4) = 16 - \frac{11}{3} = \frac{37}{3}

Final Calculation

The problem requires the value of

6 f_{1} f_{2} - r

. Substituting our derived values:

6 (\frac{37}{3}) - 4 = 2 (37) - 4 = 74 - 4 = 70

The final result is 70.

Common Traps & FAQs

Wait, why is the focal distance formula

b \pm ey

and not

a \pm e x

? I always get confused with the orientation.

That is a classic trap! The formula depends entirely on the major axis. If the major axis is along the

y

-axis (a vertical ellipse), the distance from any point

(x, y)

to the foci is

b \pm ey

. If it were a horizontal ellipse, it would be

a \pm e x

. Always check which denominator is larger: if the

y^{2}

term has the larger denominator, it is a vertical ellipse. In this problem,

b^{2} = 16

and

a^{2} = 4/3

, so it is definitely vertical!

Can I just use the distance formula from the point

(1, 2)

to the foci

(0, \pm b e)

You absolutely could, but it would be a nightmare of algebra! You would have to find the coordinates of the foci, calculate the distance to each, and then multiply them. Using the property

f_{1} f_{2} = b^{2} - e^{2} y^{2}

is the 'pro' move. It leverages the definition of the ellipse and saves you from dealing with square roots and complex coordinate geometry.

Why do we need to find

p

first? Can't we just work with the ellipse equation as is?

You cannot proceed without

p

because the ellipse is not fully defined! The problem gives you the ellipse

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

, but

p

is an unknown. Since the problem tells us the ellipse passes through the center of the circle, that point must satisfy the equation. Finding

p

is the key that unlocks the entire problem. Without it, you do not have a specific ellipse to work with.

Animated Solution for Mathematics - Conic Sections: Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre $C$ of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius $r$ . Let $f_{1}, f_{2}$ be the focal distances of the point $C$ on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to

Select Answer:

Visualized Solution

Center and Radius of the Circle

Finding the Unknown Parameter $p$

Standard Form of the Ellipse

Identifying the Major Axis

Finding the Eccentricity

Formula for Focal Distances

Computing the Product of Focal Distances

Evaluating the Final Expression

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

Analyzing the Circle

Determining the Ellipse Parameters

Geometry of the Ellipse

Calculating Focal Distances

Final Calculation

Similar Questions

The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

Let $S$ and $S^{'}$ be the foci of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If $(S P)^{2} + (S^{'} P)^{2} - S P \cdot S^{'} P = 37$ , then $α^{2} + β^{2}$ is equal to :

An ellipse is drawn by taking a diameter of the circle $(x - 1)^{2} + y^{2} = 1$ as its semi-minor axis and a diameter of the circle $x^{2} + (y - 2)^{2} = 4$ as semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is:

Select Answer:

Visualized Solution

Center and Radius of the Circle

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

Standard Form of the Ellipse

Identifying the Major Axis

Finding the Eccentricity

Formula for Focal Distances

Computing the Product of Focal Distances

Evaluating the Final Expression

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

Analyzing the Circle

Determining the Ellipse Parameters

Geometry of the Ellipse

Calculating Focal Distances

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line y=x+1 meets the ellipse 4x2​+2y2​=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to

Finding the Unknown Parameter $p$

The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to