JEE MainMathematics

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Locus of the Circumcentre of a Right-Angled Triangle Formed by Parabola Tangents

Q: Why is the circumcentre simply the midpoint of the hypotenuse?

The angle between the lines x = 0 and y = 3 is 90°. In any right-angled triangle, the hypotenuse is the diameter of the circumcircle, so the circumcentre is always the midpoint of the hypotenuse.

The equations of two sides of a variable triangle are x=0 and y=3, and its third side is a tangent to parabola y^2=6x. The locus of its circumcentre is:

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A4y^2 - 18y - 3x - 18 = 0B4y^2 - 18y - 3x + 18 = 0C4y^2 - 18y + 3x + 18 = 0D4y^2 + 18y + 3x + 18 = 0

Step-by-step solutionView

Correct answer

Since two sides are $x = 0$ and $y = 3$ , the triangle is right-angled at $(0, 3)$ , meaning its circumcentre is the midpoint of the hypotenuse lying on the tangent line $y = m x + \frac{3}{2 m}$ , yielding the locus $4 y^{2} - 18 y + 3 x + 18 = 0$ .

Option analysis

Why each option works or fails

A · 4y^2 - 18y - 3x - 18 = 0

The student made a sign error when solving for the $x$ -intercept or in the elimination of the parameter $m$ , leading to negative signs on both the $x$ and constant terms. Ensure careful tracking of signs when finding $x_{1} = - \frac{3}{2 m^{2}}$ and simplifying the parameter $m = \frac{2 y - 3}{2 h}$ .

B · 4y^2 - 18y - 3x + 18 = 0

The student incorrectly handled the sign of $x$ when substituting $m = \frac{2 k - 3}{2 h}$ into $h = - \frac{3}{4 m^{2}}$ , dropping the negative sign. The tangent $y = m x + \frac{3}{2 m}$ intersects $y = 3$ at $x = \frac{3 - 3 / (2 m)}{m}$ , and $x = 0$ at $y = \frac{3}{2 m}$ . Check the signs carefully when deriving $2 h$ .

C · 4y^2 - 18y + 3x + 18 = 0

This is the correct equation of the locus obtained by eliminating $m$ from the circumcentre coordinates. Correctly identify the triangle as right-angled at $(0, 3)$ , express the circumcentre as the midpoint of the other two vertices on the tangent, and eliminate $m$ .

D · 4y^2 + 18y + 3x + 18 = 0

The student inverted the sign inside the binomial $(2 y - 3)^{2}$ , writing it as $(2 y + 3)^{2} = 4 y^{2} + 12 y + 9$ or slipping on the middle linear term of $y$ . Since the right angle is at $(0, 3)$ , the vertical coordinate involves $(y - 3)$ and $(2 y - 3)^{2}$ , which expands to $4 y^{2} - 12 y + 9$ , not $+ 18 y$ .

Reviewed route

Solution

StepWorking

01Given

Two fixed sides of the triangle are $x = 0$ and $y = 3$ . The third side is a tangent to the parabola $y^{2} = 6 x$ . Note that $x = 0$ is perpendicular to $y = 3$ .

02Goal

Find the locus of the circumcentre $(h, k)$ of the triangle as the tangent varies.

03Approach

Since $x = 0$ and $y = 3$ are perpendicular, the triangle is right-angled at $(0, 3)$ . Therefore, its circumcentre $(h, k)$ is simply the midpoint of the hypotenuse, which lies on the tangent line between its intersections with $x = 0$ and $y = 3$ .

04Execute

For the parabola $y^{2} = 4 a x$ with $4 a = 6 ⟹ a = 3 / 2$ , the equation of tangent in parametric form is $y t = x + a t^{2} = x + \frac{3}{2} t^{2}$ , or $x - y t + \frac{3}{2} t^{2} = 0$ .

05Execute

Find intersections with the two axes/lines: 1) With $x = 0$ : $- y t + \frac{3}{2} t^{2} = 0 ⟹ y = \frac{3}{2} t$ , so $P = (0, \frac{3 t}{2})$ . 2) With $y = 3$ : $x - 3 t + \frac{3}{2} t^{2} = 0 ⟹ x = 3 t - \frac{3}{2} t^{2} = \frac{6 t - 3 t^{2}}{2}$ , so $Q = (\frac{6 t - 3 t^{2}}{2}, 3)$ .

06Execute

The circumcentre $(h, k)$ is the midpoint of $P Q$ : $h = \frac{0 + \frac{6 t - 3 t^{2}}{2}}{2} = \frac{6 t - 3 t^{2}}{4} ⟹ 4 h = 6 t - 3 t^{2}$ $k = \frac{\frac{3 t}{2} + 3}{2} = \frac{3 t + 6}{4} ⟹ 4 k - 6 = 3 t ⟹ t = \frac{4 k - 6}{3}$ .

07Execute

Substitute $t = \frac{4 k - 6}{3}$ into $4 h = 6 t - 3 t^{2}$ : $4 h = 2 (4 k - 6) - 3 {(\frac{4 k - 6}{3})}^{2} = 8 k - 12 - \frac{(4 k - 6)^{2}}{3}$ Multiply by 3: $12 h = 24 k - 36 - (16 k^{2} - 48 k + 36) = - 16 k^{2} + 72 k - 72$ Divide by 4 and rearrange: $4 k^{2} - 18 k + 3 h + 18 = 0$ . Replacing $(h, k)$ with $(x, y)$ , the locus is $4 y^{2} - 18 y + 3 x + 18 = 0$ .

✓Verify

For $t = 2$ : $k = 3$ , $t = \frac{4 (3) - 6}{3} = 2$ . Then $4 h = 6 (2) - 3 (4) = 0 ⟹ h = 0$ . Check point $(0, 3)$ in locus: $4 (9) - 18 (3) + 3 (0) + 18 = 36 - 54 + 18 = 0$ , which is satisfied.

Hints that build this answer step by step

The sides $x = 0$ and $y = 3$ are perpendicular. Where is the circumcentre of any right-angled triangle located?
At the midpoint of the hypotenuse (the third side)
A general tangent to the parabola $y^{2} = 6 x$ with slope $m$ has equation $y = m x + \frac{3}{2 m}$ . What are its intersection points with $x = 0$ and $y = 3$ ?
$A (0, \frac{3}{2 m})$ and $B (\frac{3 - \frac{3}{2 m}}{m}, 3)$
If $(h, k)$ is the midpoint of $A (0, \frac{3}{2 m})$ and $B (\frac{6 m - 3}{2 m^{2}}, 3)$ , what are the parametric equations for $h$ and $k$ ?
$2 h = \frac{6 m - 3}{2 m^{2}}$ and $2 k = 3 + \frac{3}{2 m}$
From $2 k - 3 = \frac{3}{2 m}$ , we have $\frac{1}{m} = \frac{2 (2 k - 3)}{3}$ . Substituting this into $4 h = \frac{6}{m} - \frac{3}{m^{2}}$ , what relation between $x$ and $y$ is obtained?
$4 y^{2} - 18 y + 3 x + 18 = 0$

✓ Source and academic review

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

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Students also ask

Why is the circumcentre simply the midpoint of the hypotenuse?

The angle between the lines $x = 0$ and $y = 3$ is $90^{\circ}$ . In any right-angled triangle, the hypotenuse is the diameter of the circumcircle, so the circumcentre is always the midpoint of the hypotenuse.

Answer

Since two sides are $x = 0$ and $y = 3$ , the triangle is right-angled at $(0, 3)$ , meaning its circumcentre is the midpoint of the hypotenuse lying on the tangent line $y = m x + \frac{3}{2 m}$ , yielding the locus $4 y^{2} - 18 y + 3 x + 18 = 0$ .

Why each option works or fails

A: 4y^2 - 18y - 3x - 18 = 0 - The student made a sign error when solving for the $x$ -intercept or in the elimination of the parameter $m$ , leading to negative signs on both the $x$ and constant terms. Ensure careful tracking of signs when finding $x_{1} = - \frac{3}{2 m^{2}}$ and simplifying the parameter $m = \frac{2 y - 3}{2 h}$ .
B: 4y^2 - 18y - 3x + 18 = 0 - The student incorrectly handled the sign of $x$ when substituting $m = \frac{2 k - 3}{2 h}$ into $h = - \frac{3}{4 m^{2}}$ , dropping the negative sign. The tangent $y = m x + \frac{3}{2 m}$ intersects $y = 3$ at $x = \frac{3 - 3 / (2 m)}{m}$ , and $x = 0$ at $y = \frac{3}{2 m}$ . Check the signs carefully when deriving $2 h$ .
C · correct: 4y^2 - 18y + 3x + 18 = 0 - This is the correct equation of the locus obtained by eliminating $m$ from the circumcentre coordinates. Correctly identify the triangle as right-angled at $(0, 3)$ , express the circumcentre as the midpoint of the other two vertices on the tangent, and eliminate $m$ .
D: 4y^2 + 18y + 3x + 18 = 0 - The student inverted the sign inside the binomial $(2 y - 3)^{2}$ , writing it as $(2 y + 3)^{2} = 4 y^{2} + 12 y + 9$ or slipping on the middle linear term of $y$ . Since the right angle is at $(0, 3)$ , the vertical coordinate involves $(y - 3)$ and $(2 y - 3)^{2}$ , which expands to $4 y^{2} - 12 y + 9$ , not $+ 18 y$ .

Step-by-step solution

Given: Two fixed sides of the triangle are $x = 0$ and $y = 3$ . The third side is a tangent to the parabola $y^{2} = 6 x$ . Note that $x = 0$ is perpendicular to $y = 3$ .
Goal: Find the locus of the circumcentre $(h, k)$ of the triangle as the tangent varies.
Approach: Since $x = 0$ and $y = 3$ are perpendicular, the triangle is right-angled at $(0, 3)$ . Therefore, its circumcentre $(h, k)$ is simply the midpoint of the hypotenuse, which lies on the tangent line between its intersections with $x = 0$ and $y = 3$ .
Execute: For the parabola $y^{2} = 4 a x$ with $4 a = 6 ⟹ a = 3 / 2$ , the equation of tangent in parametric form is $y t = x + a t^{2} = x + \frac{3}{2} t^{2}$ , or $x - y t + \frac{3}{2} t^{2} = 0$ .
Execute: Find intersections with the two axes/lines: 1) With $x = 0$ : $- y t + \frac{3}{2} t^{2} = 0 ⟹ y = \frac{3}{2} t$ , so $y t = x + a t^{2} = x + \frac{3}{2} t^{2}$ . 2) With $y = 3$ : $x - 3 t + \frac{3}{2} t^{2} = 0 ⟹ x = 3 t - \frac{3}{2} t^{2} = \frac{6 t - 3 t^{2}}{2}$ , so $Q = (\frac{6 t - 3 t^{2}}{2}, 3)$ .
Execute: The circumcentre $(h, k)$ is the midpoint of $P Q$ : $h = \frac{0 + \frac{6 t - 3 t^{2}}{2}}{2} = \frac{6 t - 3 t^{2}}{4} ⟹ 4 h = 6 t - 3 t^{2}$ $k = \frac{\frac{3 t}{2} + 3}{2} = \frac{3 t + 6}{4} ⟹ 4 k - 6 = 3 t ⟹ t = \frac{4 k - 6}{3}$ .
Execute: Substitute $t = \frac{4 k - 6}{3}$ into $4 h = 6 t - 3 t^{2}$ : $4 h = 2 (4 k - 6) - 3 {(\frac{4 k - 6}{3})}^{2} = 8 k - 12 - \frac{(4 k - 6)^{2}}{3}$ Multiply by 3: $12 h = 24 k - 36 - (16 k^{2} - 48 k + 36) = - 16 k^{2} + 72 k - 72$ Divide by 4 and rearrange: $4 k^{2} - 18 k + 3 h + 18 = 0$ . Replacing $(h, k)$ with $(x, y)$ , the locus is $4 y^{2} - 18 y + 3 x + 18 = 0$ .
Verify: For $t = 2$ : $k = 3$ , $t = \frac{4 (3) - 6}{3} = 2$ . Then $4 h = 6 (2) - 3 (4) = 0 ⟹ h = 0$ . Check point $(0, 3)$ in locus: $4 (9) - 18 (3) + 3 (0) + 18 = 36 - 54 + 18 = 0$ , which is satisfied.

Locus of the Circumcentre of a Right-Angled Triangle Formed by Parabola Tangents

What feels right?

Since two sides are $x = 0$ and $y = 3$ , the triangle is right-angled at $(0, 3)$ , meaning its circumcentre is the midpoint of the hypotenuse lying on the tangent line $y = m x + \frac{3}{2 m}$ , yielding the locus $4 y^{2} - 18 y + 3 x + 18 = 0$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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