MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

The equations of two sides of a variable triangle are x=0x = 0 and y=3y = 3, and its third side is a tangent to the parabola y2=6xy^2 = 6x. The locus of its circumcentre is:

  • A

    4y218y3x18=04y^2 - 18y-3x - 18 = 0

  • B

    4y2+18y+3x+18=04y^2 + 18y + 3x + 18 = 0

  • C

    4y218y+3x+18=04y^2 - 18y + 3x + 18 = 0

  • D

    4y218y3x+18=04y^2 - 18y - 3x + 18 = 0

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two sides of the triangle are x=0x = 0 and y=3y = 3. The third side is a tangent to the parabola y2=6xy^2 = 6x.

Find: The locus of the circumcentre of the triangle.

For the parabola y2=4axy^2 = 4ax, we have

4a=64a = 6

so

a=32a = \frac{3}{2}

A tangent to y2=4axy^2 = 4ax in slope form is

y=mx+amy = mx + \frac{a}{m}

Hence here the tangent is

y=mx+32my = mx + \frac{3}{2m}

with m0m \ne 0.

Coordinate figure showing lines x equals 0, y equals 3, a tangent y equals mx plus 3 by 2m, its intercepts, and circumcentre marked as h comma k.

This tangent meets x=0x = 0 at

(0,32m)\left(0, \frac{3}{2m}\right)

and meets y=3y = 3 at

(6m32m2,3)\left(\frac{6m - 3}{2m^2}, 3\right)

Since the two given sides are perpendicular, the triangle is right-angled at (0,3)\left(0,3\right). Therefore, its circumcentre is the midpoint of the hypotenuse.

So if the circumcentre is (h,k)\left(h,k\right), then

h=6m34m2,k=6m+34mh = \frac{6m - 3}{4m^2}, \qquad k = \frac{6m + 3}{4m}

Now eliminate mm using the given relation from the solution:

3h=2(2k2+9k9)3h = 2\left(-2k^2 + 9k - 9\right)

which gives

3h=4k2+18k183h = -4k^2 + 18k - 18

Therefore,

4k218k+3h+18=04k^2 - 18k + 3h + 18 = 0

Replacing (h,k)\left(h,k\right) by (x,y)\left(x,y\right), the locus is

4y218y+3x+18=04y^2 - 18y + 3x + 18 = 0

Therefore, the correct option is C.

Using midpoint of hypotenuse

Given: The triangle is formed by the lines x=0x = 0, y=3y = 3, and a tangent to y2=6xy^2 = 6x.

Find: The locus of the circumcentre.

The intersection of x=0x = 0 and y=3y = 3 is (0,3)\left(0,3\right), so the angle between these two sides is a right angle.

Let the tangent be

y=mx+32my = mx + \frac{3}{2m}

Then its intercept on x=0x = 0 is

A(0,32m)A\left(0, \frac{3}{2m}\right)

and its intercept on y=3y = 3 is found from

3=mx+32m3 = mx + \frac{3}{2m}

so

mx=332m=6m32mmx = 3 - \frac{3}{2m} = \frac{6m - 3}{2m}

and hence

x=6m32m2x = \frac{6m - 3}{2m^2}

Thus

B(6m32m2,3)B\left(\frac{6m - 3}{2m^2}, 3\right)

The hypotenuse is ABAB, so the circumcentre is its midpoint:

h=12(0+6m32m2)=6m34m2h = \frac{1}{2}\left(0 + \frac{6m - 3}{2m^2}\right) = \frac{6m - 3}{4m^2} k=12(32m+3)=6m+34mk = \frac{1}{2}\left(\frac{3}{2m} + 3\right) = \frac{6m + 3}{4m}

Eliminating mm gives

4y218y+3x+18=04y^2 - 18y + 3x + 18 = 0

So the locus of the circumcentre is the parabola represented by this equation.

Therefore, the correct option is C.

Common mistakes

  • Treating the circumcentre as the intersection of angle bisectors without using the right-triangle property is inefficient here. Since x=0x = 0 and y=3y = 3 are perpendicular, the triangle is right-angled, so the circumcentre is the midpoint of the hypotenuse.

  • Using the wrong tangent form for y2=4axy^2 = 4ax is a common error. The slope form is y=mx+amy = mx + \frac{a}{m}, not a tangent form from another conic. Here a=32a = \frac{3}{2}, so the tangent is y=mx+32my = mx + \frac{3}{2m}.

  • While finding the intersection with y=3y = 3, students often make an algebraic mistake in solving for xx. Substitute carefully into the tangent equation and isolate xx before taking the midpoint.

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