JEE Main 2024 (05 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the line $2 x + 3 y - k = 0, k > 0$ , intersect the $x$ -axis and $y$ -axis at the points $A$ and $B$ , respectively. If the equation of the circle having the line segment $A B$ as a diameter is $x^{2} + y^{2} - 3 x - 2 y = 0$ and the length of the latus rectum of the ellipse $x^{2} + 9 y^{2} = k^{2}$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $2 m + n$ is equal to

Q: Wait, why do we reject the negative root here?

Actually, the problem states $k > 0$, so we don't have to worry about negative roots! Always check the constraints given in the problem statement first. It is a common trap to solve for a variable and forget the domain restrictions.

Q: Why can't we just use the distance formula for the diameter?

You absolutely could! You could find the length of $AB$ using the distance formula and equate it to the diameter of the circle. However, using the center (midpoint) is often algebraically cleaner and less prone to calculation errors. In JEE, choosing the path of least resistance is a skill in itself.

Q: How do I know if the ellipse is horizontal or vertical?

Look at the denominators. In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, if the denominator under $x^2$ is larger than the one under $y^2$, the major axis is horizontal. Here, $36 > 4$, so it is horizontal. Always normalize the equation to $1$ on the right side first!

Select Answer:

Visualized Solution

Equation of the Line

Given line: $2 x + 3 y - k = 0$

Coordinates of Point $A$

For x-intercept, set $y = 0$
$2 x - k = 0 ⟹ x = \frac{k}{2}$
Point $A = (\frac{k}{2}, 0)$

Coordinates of Point $B$

For y-intercept, set $x = 0$
$3 y - k = 0 ⟹ y = \frac{k}{3}$
Point $B = (0, \frac{k}{3})$

Circle with Diameter $A B$

Given circle: $x^{2} + y^{2} - 3 x - 2 y = 0$
Line segment $A B$ is the diameter of this circle.

Finding the Circle's Center

General circle equation: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Center is $(- g, - f)$
For our circle, $2 g = - 3 ⟹ - g = \frac{3}{2}$
$2 f = - 2 ⟹ - f = 1$
Center $C = (\frac{3}{2}, 1)$

Center as Midpoint of $A B$

Since $A B$ is the diameter, its midpoint is the center $C$ .
Midpoint of $A (\frac{k}{2}, 0)$ and $B (0, \frac{k}{3})$ :
$(\frac{\frac{k}{2} + 0}{2}, \frac{0 + \frac{k}{3}}{2}) = (\frac{k}{4}, \frac{k}{6})$

Equating Coordinates to Find $k$

Equate x-coordinates: $\frac{k}{4} = \frac{3}{2}$
$⟹ k = 6$
Check y-coordinates: $\frac{k}{6} = 1 ⟹ k = 6$

The Ellipse Equation

Given ellipse: $x^{2} + 9 y^{2} = k^{2}$
Substitute $k = 6$ : $x^{2} + 9 y^{2} = 36$

Standard Form of the Ellipse

Divide by $36$ : $\frac{x ^{2}}{36} + \frac{9 y ^{2}}{36} = 1$
$⟹ \frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1$
Compare with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
$a^{2} = 36 ⟹ a = 6$
$b^{2} = 4 ⟹ b = 2$

Length of Latus Rectum

Formula for length of Latus Rectum ( $L R$ ): $\frac{2 b ^{2}}{a}$
Substitute $a = 6$ and $b^{2} = 4$ :

Calculating $L R$

$L R = \frac{2 ( 4 )}{6}$
$L R = \frac{8}{6} = \frac{4}{3}$

Comparing with $\frac{m}{n}$

Given $L R = \frac{m}{n}$
We found $L R = \frac{4}{3}$
Since $4$ and $3$ are coprime (no common factors), $m = 4$ and $n = 3$ .

Final Value: $2 m + n$

We need to find the value of $2 m + n$
Substitute $m = 4, n = 3$ :
$2 (4) + 3 = 8 + 3 = 11$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The line is given by the equation

2 x + 3 y - k = 0

. This line acts as our anchor, creating intercepts

A

and

B

on the coordinate axes.

To find the

x

-intercept, we set

y = 0

, which yields

2 x = k

, or

x = \frac{k}{2}

. For the

y

-intercept, we set

x = 0

, which yields

3 y = k

, or

y = \frac{k}{3}

Thus, our points are defined as

A = (\frac{k}{2}, 0)

and

B = (0, \frac{k}{3})

The Circle's Secret

We are given the circle equation

x^{2} + y^{2} - 3 x - 2 y = 0

. By comparing this to the general form

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

, we identify the center

(- g, - f)

For our equation,

2 g = - 3

and

2 f = - 2

, which gives us the center

C = (\frac{3}{2}, 1)

Geometrically, the center of a circle is the midpoint of its diameter

A B

. We calculate the midpoint of

A B

using the coordinates of

A

and

B

M i d p o in t = (\frac{\frac{k}{2} + 0}{2}, \frac{0 + \frac{k}{3}}{2}) = (\frac{k}{4}, \frac{k}{6})

By equating the midpoint

(\frac{k}{4}, \frac{k}{6})

to the center

(\frac{3}{2}, 1)

, we solve for

k

\frac{k}{4} = \frac{3}{2} \Rightarrow k = 6

The Ellipse and the Latus Rectum

With

k = 6

, the ellipse equation

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

becomes

x^{2} + 9 y^{2} = 36

. Dividing by

36

to reach the standard form, we obtain:

\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1

This is a horizontal ellipse where

a^{2} = 36

(so

a = 6

) and

b^{2} = 4

(so

b = 2

). The length of the latus rectum is given by the formula:

L R = \frac{2 b ^{2}}{a}

Substituting our values into the formula:

L R = \frac{2 ( 4 )}{6} = \frac{8}{6} = \frac{4}{3}

Final Calculation

We are given that the length of the latus rectum is

\frac{m}{n}

, where

m

and

n

are coprime. Therefore,

m = 4

and

n = 3

The final step is to calculate

2 m + n

2 (4) + 3 = 8 + 3 = 11

The final answer is 11.

Common Traps & FAQs

Wait, why do we reject the negative root here?

Actually, the problem states

k > 0

, so we don't have to worry about negative roots! Always check the constraints given in the problem statement first. It is a common trap to solve for a variable and forget the domain restrictions.

Why can't we just use the distance formula for the diameter?

You absolutely could! You could find the length of

A B

using the distance formula and equate it to the diameter of the circle. However, using the center (midpoint) is often algebraically cleaner and less prone to calculation errors. In JEE, choosing the path of least resistance is a skill in itself.

How do I know if the ellipse is horizontal or vertical?

Look at the denominators. In the standard form

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

, if the denominator under

x^{2}

is larger than the one under

y^{2}

, the major axis is horizontal. Here,

36 > 4

, so it is horizontal. Always normalize the equation to

1

on the right side first!

Select Answer:

Visualized Solution

Equation of the Line

Coordinates of Point $A$

Coordinates of Point $B$

Circle with Diameter $A B$

Finding the Circle's Center

Center as Midpoint of $A B$

Equating Coordinates to Find $k$

The Ellipse Equation

Standard Form of the Ellipse

Length of Latus Rectum

Calculating $L R$

Comparing with $\frac{m}{n}$

Final Value: $2 m + n$

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

Analyzing the Setup

The Circle's Secret

The Ellipse and the Latus Rectum

Final Calculation

Similar Questions

Let the latus rectum of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + n)$ , where $l, m$ and $n$ are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to ______

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

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , then the length of its latus rectum is:

Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be an ellipse, whose eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $14$ . Then the square of the eccentricity of $\frac{x ^{2}}{a ^{2}} - \frac{m y ^{2}}{b ^{2}} = 1$ is :

Let the length of a latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be 10 . If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to :

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2 x + y = 6$ . Let an ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$ . Then the square of the length of the latus rectum of $E$ , is

Select Answer:

Visualized Solution

Equation of the Line

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

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

Circle with Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Finding the Circle's Center

Center as Midpoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Equating Coordinates to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

The Ellipse Equation

Standard Form of the Ellipse

Length of Latus Rectum

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

Comparing with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nm​

Final Value: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2m+n

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

Analyzing the Setup

The Circle's Secret

The Ellipse and the Latus Rectum

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

Coordinates of Point $A$

Coordinates of Point $B$

Circle with Diameter $A B$

Center as Midpoint of $A B$

Equating Coordinates to Find $k$

Calculating $L R$

Comparing with $\frac{m}{n}$

Final Value: $2 m + n$

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