JEE Main 2024 (01 Feb Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: Let the line $L : 2 x + y = α$ pass through the point of the intersection $P$ (in the first quadrant) of the circle $x^{2} + y^{2} = 3$ and the parabola $x^{2} = 2 y$ . Let the line $L$ touch two circles $C_{1}$ and $C_{2}$ of equal radius $23$ . If the centres $Q_{1}$ and $Q_{2}$ of the circles $C_{1}$ and $C_{2}$ lie on the y-axis, then the square of the area of the triangle $P Q_{1} Q_{2}$ is equal to

Q: Wait, why do we reject the negative root when solving $y^2 + 2y - 3 = 0$?

Great catch! The problem explicitly states that point $P$ lies in the first quadrant. In the first quadrant, both $x$ and $y$ must be positive. Since $y = -3$ would place the point outside the first quadrant, we discard it to stay true to the geometric constraints of the problem.

Q: Why can't we just use the distance between the centers to find the area?

You absolutely can! The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. If you treat the segment $Q_1Q_2$ as the base, its length is simply the difference in their y-coordinates. The height of the triangle is then the perpendicular distance from $P$ to the y-axis, which is just the x-coordinate of $P$. It is a beautiful, simple way to finish the problem.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Before you dive into the geometry, visualize the intersection. You have a circle and a parabola meeting in the first quadrant. The point

P

is the anchor of this entire problem. Find it first by substituting

x^{2} = 2 y

into the circle equation

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

Hint 2

Hint 2: The Arsenal. Once you have

P

, you define the line

L

. The key to finding the centers

Q_{1}

and

Q_{2}

is the geometric definition of a tangent. Remember that the perpendicular distance from the center of a circle to its tangent line is exactly equal to the radius

r = 23

Hint 3

Hint 3: The Execution. Set the centers as

(0, k)

since they lie on the y-axis. Use the distance formula

\frac{∣ 2 ( 0 ) + k - 3∣}{( 2 ) ^{2} + 1 ^{2}} = 23

to solve for

k

. This will give you two distinct values, defining the two possible centers.

Visualized Solution

Visualizing the Curves

Given Circle: $x^{2} + y^{2} = 3$
Given Parabola: $x^{2} = 2 y$
Point $P$ is the intersection in the first quadrant.

Setting up the Intersection

Substitute $x^{2} = 2 y$ into $x^{2} + y^{2} = 3$ :
$(2 y) + y^{2} = 3 \Rightarrow y^{2} + 2 y - 3 = 0$

Solving for $y$

Factorizing: $(y + 3) (y - 1) = 0$
Since $P$ is in the first quadrant, $y > 0 \Rightarrow y = 1$

Finding Point $P$

Substitute $y = 1$ back: $x^{2} = 2 (1) \Rightarrow x = 2$
Point $P = (2, 1)$

Equation of Line $L$

Line $L : 2 x + y = α$ passes through $P (2, 1)$
Substitute coordinates: $2 (2) + 1 = α$

Finding $α$

Calculation: $2 + 1 = α \Rightarrow α = 3$
Equation of Line $L : 2 x + y - 3 = 0$

Locating Centers $Q_{1}$ and $Q_{2}$

Centers $Q_{1}, Q_{2}$ lie on y-axis: Let them be $(0, k)$
Radius $r = 23$

Distance Formula Setup

Distance from $(0, k)$ to $2 x + y - 3 = 0$ is $r$ :
Formula: $\frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}} = r$

Substituting into Distance Formula

Substitute values:
$\frac{∣ 2 ( 0 ) + k - 3∣}{( 2 ) ^{2} + 1 ^{2}} = 23$

Simplifying the Equation

Simplify denominator: $2 + 1 = 3$
$\frac{∣ k - 3∣}{3} = 23 \Rightarrow ∣ k - 3∣ = 6$

Solving for $k$

Case 1: $k - 3 = 6 \Rightarrow k = 9$
Case 2: $k - 3 = - 6 \Rightarrow k = - 3$
Centers: $Q_{1} (0, 9)$ and $Q_{2} (0, - 3)$

Forming Triangle $P Q_{1} Q_{2}$

Vertices of $△ P Q_{1} Q_{2}$ : $P (2, 1)$ , $Q_{1} (0, 9)$ , $Q_{2} (0, - 3)$

Calculating Area

Base $Q_{1} Q_{2} = ∣9 - (- 3) ∣ = 12$
Height (x-coordinate of $P$ ) $= 2$
Area $= \frac{1}{2} \times 12 \times 2 = 62$

Final Answer

Area of $△ P Q_{1} Q_{2} = 62$
Square of the Area $= (62)^{2}$
Calculation: $36 \times 2 = 72$
Final Answer: 72

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Finding the Anchor

Point

P

To find

P

, we look for the intersection of the circle

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

and the parabola

x^{2} = 2 y

. By substituting the parabola's equation into the circle's equation, we transform the system into a quadratic in terms of

y

2 y + y^{2} = 3

Rearranging this gives us the quadratic equation

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

. Factoring this expression, we obtain:

(y + 3) (y - 1) = 0

Since we are restricted to the first quadrant,

y

must be positive, which implies

y = 1

. Plugging this value back into the parabola equation

x^{2} = 2 (1)

, we find

x = 2

. Thus, our anchor point is

P (2, 1)

The Line of Tangency

Now that we have

P

, we define our line

L : 2 x + y = α

. Since

L

passes through

P (2, 1)

, we substitute these coordinates into the equation:

2 (2) + 1 = α ⟹ α = 3

Our line is now fully revealed as

2 x + y - 3 = 0

. This line serves as the geometric constraint for our circles

C_{1}

and

C_{2}

The Geometry of Centers

We are given that the centers

Q_{1}

and

Q_{2}

lie on the

y

-axis. Let their coordinates be

(0, k)

. The radius of these circles is

r = 23

The condition for tangency dictates that the perpendicular distance from the center

(0, k)

to the line

2 x + y - 3 = 0

must equal the radius. Using the distance formula, we write:

\frac{∣ 2 ( 0 ) + k - 3∣}{( 2 ) ^{2} + 1 ^{2}} = 23

Simplifying the denominator, we get

3

. Multiplying both sides by

3

, we arrive at

∣ k - 3∣ = 6

. This yields two possible values for

k

k = 9

and

k = - 3

. Consequently, our centers are

Q_{1} (0, 9)

and

Q_{2} (0, - 3)

The Final Act

The Triangle

We now calculate the area of

△ P Q_{1} Q_{2}

with vertices

P (2, 1)

Q_{1} (0, 9)

, and

Q_{2} (0, - 3)

. The base of this triangle lies on the

y

-axis, with length:

Q_{1} Q_{2} = ∣9 - (- 3) ∣ = 12

The height of the triangle is the perpendicular distance from

P

to the

y

-axis, which is simply the

x

-coordinate of

P

, namely

2

. The area is given by:

Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 12 \times 2 = 62

The problem asks for the square of the area. Squaring

62

, we get:

(62)^{2} = 36 \times 2 = 72

Common Traps & FAQs

Wait, why do we reject the negative root when solving

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

Great catch! The problem explicitly states that point

P

lies in the first quadrant. In the first quadrant, both

x

and

y

must be positive. Since

y = - 3

would place the point outside the first quadrant, we discard it to stay true to the geometric constraints of the problem.

Why can't we just use the distance between the centers to find the area?

You absolutely can! The area of a triangle is

\frac{1}{2} \times base \times height

. If you treat the segment

Q_{1} Q_{2}

as the base, its length is simply the difference in their y-coordinates. The height of the triangle is then the perpendicular distance from

P

to the y-axis, which is just the x-coordinate of

P

. It is a beautiful, simple way to finish the problem.

Is it possible for the line

L

to be tangent to only one circle?

In this specific configuration, the math forces two solutions for

k

because the line

L

is fixed, and there are two points on the y-axis that are exactly

23

distance away from that line. Both are valid centers for circles of the given radius.

Do I need to worry about the orientation of the triangle?

Not at all! The area of a triangle is a scalar quantity. Whether the triangle is 'tilted' or 'upside down' doesn't change the magnitude of the area. As long as you correctly identify the base and the perpendicular height, the formula will yield the correct result.

Why is the final answer the square of the area?

That is a classic JEE twist! Often, examiners ask for the square of the area to ensure that if your result involves a square root (like

62

), you perform the final squaring step correctly, testing both your geometry and your algebraic precision.

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Setting up the Intersection

Solving for $y$

Finding Point $P$

Equation of Line $L$

Finding $α$

Locating Centers $Q_{1}$ and $Q_{2}$

Distance Formula Setup

Substituting into Distance Formula

Simplifying the Equation

Solving for $k$

Forming Triangle $P Q_{1} Q_{2}$

Calculating Area

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Finding the Anchor

The Line of Tangency

The Geometry of Centers

The Final Act

Similar Questions

If the tangent at a point P on the parabola $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ , then the area of the triangle PQR is:

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Setting up the Intersection

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

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

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

Locating Centers .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q2​

Distance Formula Setup

Substituting into Distance Formula

Simplifying the Equation

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

Forming Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ1​Q2​

Calculating Area

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Finding the Anchor

The Line of Tangency

The Geometry of Centers

The Final Act

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the tangents at two points A and B on the circle x2+y2−4x+3=0 meet at origin O(0,0). Then the area of the triangle of OAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangents at the point A(1,3) and B(1,−1) on the parabola y2−2x−2y=1 meet at the point P. Then the area (in unit2) of the triangle PAB is :-

Solving for $y$

Finding Point $P$

Equation of Line $L$

Finding $α$

Locating Centers $Q_{1}$ and $Q_{2}$

Solving for $k$

Forming Triangle $P Q_{1} Q_{2}$

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-