JEE MainMathematics

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Area of a Triangle Formed by a Parabola Tangent and Coordinate Lines

Q: Why is the x-intercept at -b?

The tangent equation is cy = a(x + b). Setting y = 0 forces a(x + b) = 0. Since a ∈ ℕ, a ≠ 0, which requires x + b = 0, giving x = -b.

Let S be the set of all a ∈ ℕ such that the area of the triangle formed by the tangent at the point P(b, c), b, c ∈ ℕ, on the parabola y^2=2 ax and the lines x=b, y=0 is 16 unit^2, then sum of a is equal to

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Step-by-step solutionView

Correct answer

The sum of all possible natural values of $a$ is 146.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Parabola $y^{2} = 2 a x$ with $a \in N$ . Point $P (b, c)$ lies on it with $b, c \in N$ . The triangle is bounded by the tangent at $P$ , the line $x = b$ , and the line $y = 0$ (x-axis), with area equal to $16$ .

02Goal

Find the sum of all possible natural values of $a$ , denoted $\sum_{a \in S} a$ .

03Approach

Write the equation of the tangent at $P (b, c)$ using $T = 0$ , find its x-intercept to determine the base along the x-axis, identify the vertical height corresponding to $x = b$ , equate the triangle area to 16, and use $c^{2} = 2 a b$ along with natural number constraints on $b$ to find all valid $a$ .

04Execute

The tangent at $P (b, c)$ to $y^{2} = 2 a x$ is given by $y y_{1} = a (x + x_{1}) ⟹ c y = a (x + b)$ . For the intersection with $y = 0$ (the x-axis): $0 = a (x + b) ⟹ x = - b$ . The line $x = b$ intersects $y = 0$ at $(b, 0)$ and the tangent at $P (b, c)$ . Thus, the base of the right-angled triangle along $y = 0$ spans from $x = - b$ to $x = b$ , giving length $base = b - (- b) = 2 b$ . The height of the triangle at $x = b$ is the y-coordinate of $P$ , which is $c$ . Therefore, the area of the triangle is $\frac{1}{2} \times base \times height = \frac{1}{2} \times (2 b) \times c = b c$ . Equating to the given area: $b c = 16$ .

05Execute

Since $P (b, c)$ lies on $y^{2} = 2 a x$ , we have $c^{2} = 2 a b$ . From $b c = 16$ , we have $c = \frac{16}{b}$ . Substitute $c$ into the parabola equation: ${(\frac{16}{b})}^{2} = 2 a b ⟹ \frac{256}{b^{2}} = 2 a b ⟹ a b^{3} = 128$ . Hence, $a = \frac{128}{b^{3}}$ .

06Execute

Since $a, b \in N$ , $b^{3}$ must be a factor of $128 = 2^{7}$ . The positive integer cubes dividing 128 are: - $b = 1 ⟹ b^{3} = 1 ⟹ a = 128$ (here $c = 16 \in N$ ) - $b = 2 ⟹ b^{3} = 8 ⟹ a = \frac{128}{8} = 16$ (here $c = 8 \in N$ ) - $b = 4 ⟹ b^{3} = 64 ⟹ a = \frac{128}{64} = 2$ (here $c = 4 \in N$ ) No other natural number cube divides 128. Thus, the set $S = {2, 16, 128}$ . The sum is $2 + 16 + 128 = 146$ .

✓Verify

Check $a = 2$ : $b = 4, c = 4$ . Point $P (4, 4)$ on $y^{2} = 4 x$ ( $4^{2} = 4 (4) = 16$ , valid). Base along x-axis from $- 4$ to $4$ has length $8$ , height is $4$ . Area $= \frac{1}{2} \times 8 \times 4 = 16$ . All conditions satisfied.

Hints that build this answer step by step

What are the vertices of the triangle formed by the tangent at $P (b, c)$ , the line $x = b$ , and the line $y = 0$ ?
$(- b, 0)$ , $(b, 0)$ , and $(b, c)$
Given that the area of this right triangle is 16, what relation must $b$ and $c$ satisfy?
$b c = 16$
Using $a = \frac{c^{2}}{2 b}$ and $b c = 16$ with $a, b, c \in N$ , which values of $a$ belong to $S$ ?
$a \in {2, 16, 128}$
What is the sum of the elements in $S$ ?
$146$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the x-intercept at

- b

The tangent equation is $c y = a (x + b)$ . Setting $y = 0$ forces $a (x + b) = 0$ . Since $a \in N$ , $a \neq 0$ , which requires $x + b = 0$ , giving $x = - b$ .

Answer

Accepted answer: 146

The sum of all possible natural values of $a$ is 146.

Why each option works or fails

Step-by-step solution

Given: Parabola $y^{2} = 2 a x$ with $a \in N$ . Point $P (b, c)$ lies on it with $b, c \in N$ . The triangle is bounded by the tangent at $P$ , the line $x = b$ , and the line $y = 0$ (x-axis), with area equal to $16$ .
Goal: Find the sum of all possible natural values of $a$ , denoted $\sum_{a \in S} a$ .
Approach: Write the equation of the tangent at $P (b, c)$ using $T = 0$ , find its x-intercept to determine the base along the x-axis, identify the vertical height corresponding to $x = b$ , equate the triangle area to 16, and use $c^{2} = 2 a b$ along with natural number constraints on $b$ to find all valid $a$ .
Execute: The tangent at $P (b, c)$ to $y^{2} = 2 a x$ is given by $y y_{1} = a (x + x_{1}) ⟹ c y = a (x + b)$ . For the intersection with $y = 0$ (the x-axis): $0 = a (x + b) ⟹ x = - b$ . The line $a \in N$ intersects $P (b, c)$ at $(b, 0)$ and the tangent at $P (b, c)$ . Thus, the base of the right-angled triangle along $y = 0$ spans from $x = - b$ to $x = b$ , giving length $base = b - (- b) = 2 b$ . The height of the triangle at $x = b$ is the y-coordinate of $P$ , which is $c$ . Therefore, the area of the triangle is $\frac{1}{2} \times base \times height = \frac{1}{2} \times (2 b) \times c = b c$ . Equating to the given area: $b c = 16$ .
Execute: Since $P (b, c)$ lies on $y^{2} = 2 a x$ , we have $c^{2} = 2 a b$ . From $b c = 16$ , we have $c = \frac{16}{b}$ . Substitute $c$ into the parabola equation: ${(\frac{16}{b})}^{2} = 2 a b ⟹ \frac{256}{b^{2}} = 2 a b ⟹ a b^{3} = 128$ . Hence, $a = \frac{128}{b^{3}}$ .
Execute: Since $a, b \in N$ , $b^{3}$ must be a factor of $128 = 2^{7}$ . The positive integer cubes dividing 128 are: - $b = 1 ⟹ b^{3} = 1 ⟹ a = 128$ (here $c = 16 \in N$ ) - $b = 2 ⟹ b^{3} = 8 ⟹ a = \frac{128}{8} = 16$ (here $c = 8 \in N$ ) - $b = 4 ⟹ b^{3} = 64 ⟹ a = \frac{128}{64} = 2$ (here $c = 4 \in N$ ) No other natural number cube divides 128. Thus, the set $S = {2, 16, 128}$ . The sum is $2 + 16 + 128 = 146$ .
Verify: Check $a = 2$ : $b = 4, c = 4$ . Point $P (4, 4)$ on $y^{2} = 4 x$ ( $4^{2} = 4 (4) = 16$ , valid). Base along x-axis from $- 4$ to $4$ has length $8$ , height is $4$ . Area $= \frac{1}{2} \times 8 \times 4 = 16$ . All conditions satisfied.

Area of a Triangle Formed by a Parabola Tangent and Coordinate Lines

What feels right?

The sum of all possible natural values of $a$ is 146.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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