JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let S be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P (b, c), b, c \in N$ , on the parabola $y^{2} = 2 a x$ and the lines $x = b, y = 0$ is 16 unit $^{2}$ , then $\sum_{a \in S} a$ is equal to

Q: Why is the intersection with the x-axis at $(-b, 0)$?

Great observation! If you set $y=0$ in the tangent equation $yc = a(x + b)$, you get $0 = a(x + b)$. Since $a$ is a parameter of the parabola ($a \neq 0$), we must have $x + b = 0$, which gives $x = -b$. It is a beautiful property of parabolas that the tangent at any point $P$ intersects the axis of the parabola at a distance equal to the x-coordinate of $P$ in the opposite direction.

Q: How do I know I haven't missed any values of $a$?

By setting $bc = 16$, we restricted our search to the factors of 16. Since $b$ and $c$ must be natural numbers, there are only a finite number of pairs $(b, c)$ to check. By systematically testing each factor pair, we ensure that no valid integer $a$ is left behind.

Q: Does the parabola $y^2 = 2ax$ change the way we calculate the tangent?

Not at all! The standard form is $y^2 = 4Ax$, where $4A$ is the coefficient. Here, our coefficient is $2a$, so our $A$ is effectively $a/2$. The formula $yy_1 = 2A(x + x_1)$ becomes $yy_1 = 2(a/2)(x + x_1)$, which simplifies perfectly to $yy_1 = a(x + x_1)$. The logic remains robust regardless of the coefficient.

Q: Why did we reject the cases where $a$ was a fraction?

The problem explicitly states that $a \in \mathbb{N}$, meaning $a$ must be a natural number (a positive integer). If our calculation yields $a = 1/4$ or $a = 1/32$, these values violate the initial condition of the set $S$. Always keep an eye on the domain constraints provided in the question!

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Point $P$

Given parabola: $y^{2} = 2 a x$ , where $a \in N$ .
Point $P (b, c)$ lies on the parabola, with $b, c \in N$ .
Constraint: $c^{2} = 2 ab$ .

Equation of the Tangent at $P (b, c)$

Equation of tangent at $(x_{1}, y_{1})$ to $y^{2} = 2 a x$ is $y y_{1} = a (x + x_{1})$ .
Substituting $P (b, c)$ , the tangent equation is: $y c = a (x + b)$ .
Intersection with $y = 0$ (x-axis): $0 = a (x + b) ⟹ x = - b$ .
Vertex $A = (- b, 0)$ .

Identifying the Triangle Vertices

Lines forming the triangle: Tangent $y c = a (x + b)$ , Line $x = b$ , and Line $y = 0$ .
Intersection of $x = b$ and $y = 0$ : $B (b, 0)$ .
The vertices of the triangle are $A (- b, 0)$ , $B (b, 0)$ , and $P (b, c)$ .

Calculating the Area of Triangle $A B P$

Base of triangle $A B = b - (- b) = 2 b$ .
Height of triangle $B P = c - 0 = c$ .
Area $= \frac{1}{2} \times Base \times Height = \frac{1}{2} \times (2 b) \times c = b c$ .

Constraint Analysis for $b$ and $c$

Given Area $= 16 ⟹ b c = 16$ .
Since $b, c \in N$ , possible $(b, c)$ pairs are factors of $16$ .
Pairs: $(1, 16), (2, 8), (4, 4), (8, 2), (16, 1)$ .

Isolating $a$ from the Parabola Equation

From the initial constraint $c^{2} = 2 ab$ , we can solve for $a$ .
$a = \frac{c ^{2}}{2 b}$ .
We must check which $(b, c)$ pairs yield $a \in N$ .

Testing Pairs (1, 16) and (2, 8)

Case 1: $(1, 16) ⟹ a = \frac{1 6 ^{2}}{2 ( 1 )} = \frac{256}{2} = 128 \in N$ .
Case 2: $(2, 8) ⟹ a = \frac{8 ^{2}}{2 ( 2 )} = \frac{64}{4} = 16 \in N$ .

Testing Pair (4, 4)

Case 3: $(4, 4) ⟹ a = \frac{4 ^{2}}{2 ( 4 )} = \frac{16}{8} = 2 \in N$ .
We have found three valid values for $a$ so far.

Testing Pairs (8, 2) and (16, 1)

Case 4: $(8, 2) ⟹ a = \frac{2 ^{2}}{2 ( 8 )} = \frac{4}{16} = \frac{1}{4} \in / N$ .
Case 5: $(16, 1) ⟹ a = \frac{1 ^{2}}{2 ( 16 )} = \frac{1}{32} \in / N$ .
These pairs are rejected.

Final Summation

The set of valid values is $S = {128, 16, 2}$ .
Sum of elements in $S = 128 + 16 + 2$ .
Final Sum $= 146$ .

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are exploring the parabola defined by the equation

y^{2} = 2 a x

. We consider a point

P (b, c)

lying on this curve.

Since

P

satisfies the parabola's equation, we establish our first fundamental constraint:

c^{2} = 2 ab

The Tangent as a Bridge

The equation of the tangent to the parabola

y^{2} = 2 a x

at point

P (b, c)

is given by

y c = a (x + b)

. To find where this tangent intersects the

x

-axis, we set

y = 0

This yields

0 = a (x + b)

, which implies

x = - b

. Thus, the tangent intersects the

x

-axis at the point

A (- b, 0)

Constructing the Triangle

We define a triangle with vertices at

A (- b, 0)

B (b, 0)

, and

P (b, c)

. The base of this triangle lies on the

x

-axis between

x = - b

and

x = b

, giving a length of

2 b

The height of the triangle is the vertical distance from the

x

-axis to point

P

, which is

c

. The area of this triangle is given as 16:

Area = \frac{1}{2} \times (2 b) \times c = b c = 16

The Hunt for Integers

We now have a system of two equations:

b c = 16

and

c^{2} = 2 ab

. We aim to express

a

in terms of

b

to find integer solutions.

From

b c = 16

, we have

c = \frac{16}{b}

. Substituting this into the parabola equation:

a = \frac{c ^{2}}{2 b} = \frac{( \frac{16}{b} ) ^{2}}{2 b} = \frac{256}{2 b ^{3}} = \frac{128}{b ^{3}}

For

a

to be a natural number,

b^{3}

must be a divisor of 128. We test the possible values for

b

If $b = 1$ , then $a = \frac{128}{1} = 128$ . If

b = 2

, then

a = \frac{128}{8} = 16

. If $b = 4$ , then $a = \frac{128}{64} = 2$ . If

b = 8

, then

a = \frac{128}{512} = \frac{1}{4}

(not an integer).

The Final Synthesis

The set of valid natural numbers for

a

S = {128, 16, 2}

The sum of these values is:

128 + 16 + 2 = 146

The final result is 146.

Common Traps & FAQs

Wait, why do we use the tangent formula

y y_{1} = a (x + x_{1})

instead of finding the slope first?

You absolutely could find the slope using implicit differentiation, but the formula

y y_{1} = a (x + x_{1})

is a powerful shortcut derived from the general equation of a conic. In competitive exams like JEE, speed is a virtue. Using the standard form saves you precious minutes and reduces the chance of algebraic errors.

Why is the intersection with the x-axis at

(- b, 0)

Great observation! If you set

y = 0

in the tangent equation

y c = a (x + b)

, you get

0 = a (x + b)

. Since

a

is a parameter of the parabola (

a \neq = 0

), we must have

x + b = 0

, which gives

x = - b

. It is a beautiful property of parabolas that the tangent at any point

P

intersects the axis of the parabola at a distance equal to the x-coordinate of

P

in the opposite direction.

How do I know I haven't missed any values of

a

By setting

b c = 16

, we restricted our search to the factors of 16. Since

b

and

c

must be natural numbers, there are only a finite number of pairs

(b, c)

to check. By systematically testing each factor pair, we ensure that no valid integer

a

is left behind.

Does the parabola

y^{2} = 2 a x

change the way we calculate the tangent?

Not at all! The standard form is

y^{2} = 4 A x

, where

4 A

is the coefficient. Here, our coefficient is

2 a

, so our

A

is effectively

a /2

. The formula

y y_{1} = 2 A (x + x_{1})

becomes

y y_{1} = 2 (a /2) (x + x_{1})

, which simplifies perfectly to

y y_{1} = a (x + x_{1})

. The logic remains robust regardless of the coefficient.

Why did we reject the cases where

a

was a fraction?

The problem explicitly states that

a \in N

, meaning

a

must be a natural number (a positive integer). If our calculation yields

a = 1/4

a = 1/32

, these values violate the initial condition of the set

S

. Always keep an eye on the domain constraints provided in the question!

Animated Solution for Mathematics - Conic Sections: Let S be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P (b, c), b, c \in N$ , on the parabola $y^{2} = 2 a x$ and the lines $x = b, y = 0$ is 16 unit $^{2}$ , then $\sum_{a \in S} a$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Point $P$

Equation of the Tangent at $P (b, c)$

Identifying the Triangle Vertices

Calculating the Area of Triangle $A B P$

Constraint Analysis for $b$ and $c$

Isolating $a$ from the Parabola Equation

Testing Pairs (1, 16) and (2, 8)

Testing Pair (4, 4)

Testing Pairs (8, 2) and (16, 1)

Final Summation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Bridge

Constructing the Triangle

The Hunt for Integers

The Final Synthesis

Similar Questions

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:

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$ -axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O (0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$ , then the area (in sq. units) of $Δ S O R$ is equal to

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 :-

Points $A, B$ and $C$ lie on the parabola $y^{2} = 4 a x$ . The tangents to the parabola at $A, B$ and $C$ , taken in pairs, intersect at points $P, Q$ and $R$ . Determine the ratio of the areas of the triangles $A B C$ and $P QR$ .

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to

Enter Numerical Value:

Visualized Solution

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

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(b,c)

Identifying the Triangle Vertices

Calculating the Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABP

Constraint Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a from the Parabola Equation

Testing Pairs (1, 16) and (2, 8)

Testing Pair (4, 4)

Testing Pairs (8, 2) and (16, 1)

Final Summation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Bridge

Constructing the Triangle

The Hunt for Integers

The Final Synthesis

Similar Questions

.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 :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a tangent to the Curve 9x2+16y2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

Visualizing the Parabola and Point $P$

Equation of the Tangent at $P (b, c)$

Calculating the Area of Triangle $A B P$

Constraint Analysis for $b$ and $c$

Isolating $a$ from the Parabola Equation

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 :-

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.