MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2024 Question with Solution

Let C be the circle of minimum area touching the parabola y=6x2y = 6 - x^2 and the lines y=3xy = \sqrt{3}|x|. Then, which one of the following points lies on the circle C?

  • A

    (2,4)(2, 4)

  • B

    (1,2)(1, 2)

  • C

    (2,2)(2, 2)

  • D

    (1,1)(1, 1)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Sketch showing the parabola y = 6 - x squared, the lines y = root 3 x and y = minus root 3 x, and a circle centered at (0, 6-r) tangent to them.

Given: The circle of minimum area touches the parabola y=6x2y = 6 - x^2 and the lines y=3xy = \sqrt{3}|x|.

Find: Which given point lies on the circle.

By symmetry, the center of the circle lies on the yy-axis. If the radius is rr, then the circle touches the parabola at its top side, so its center is (0,6r)\left(0, 6-r\right).

Hence the equation of the circle is

x2+(y(6r))2=r2x^2 + \left(y - (6-r)\right)^2 = r^2

The perpendicular distance from the center (0,6r)\left(0, 6-r\right) to the line y=3xy = \sqrt{3}x must be equal to the radius rr.

So,

0(6r)12+(3)2=r\frac{|0-(6-r)|}{\sqrt{1^2 + (\sqrt{3})^2}} = r

This gives

6r2=r\frac{|6-r|}{2} = r

Now solve:

6r=2r6=3rr=26-r = 2r \Rightarrow 6 = 3r \Rightarrow r = 2

The other case,

6r=2r6=rr=66-r = -2r \Rightarrow 6 = -r \Rightarrow r = -6

is not valid because radius is positive.

Therefore, r=2r = 2 and the center is (0,4)\left(0, 4\right).

So the circle is

x2+(y4)2=4x^2 + (y-4)^2 = 4

Now check the options. For (2,4)\left(2,4\right),

22+(44)2=42^2 + (4-4)^2 = 4 4+0=44 + 0 = 4

So (2,4)\left(2,4\right) lies on the circle.

Therefore, the correct option is A.

Geometric Symmetry View

Given: The parabola y=6x2y = 6 - x^2 and the pair of lines y=3xy = \sqrt{3}|x| are symmetric about the yy-axis.

Find: The point that lies on the smallest tangent circle.

Because the figure is symmetric, the required circle also has its center on the yy-axis. Let its center be (0,6r)\left(0, 6-r\right) and radius be rr. The top contact with the parabola occurs at the highest point of the circle, matching the vertical placement used in the solution.

Tangency with each line means the distance from the center to either line equals rr. Using y=3xy = \sqrt{3}x,

6r2=r\frac{|6-r|}{2} = r

This gives the only positive solution r=2r = 2.

Hence the circle becomes

x2+(y4)2=4x^2 + (y-4)^2 = 4

Testing the points:

  • (2,4):4+0=4(2,4): 4 + 0 = 4, so it lies on the circle.
  • (1,2):1+4=5(1,2): 1 + 4 = 5, so it does not lie on the circle.
  • (2,2):4+4=8(2,2): 4 + 4 = 8, so it does not lie on the circle.
  • (1,1):1+9=10(1,1): 1 + 9 = 10, so it does not lie on the circle.

Therefore, the point on the circle is (2,4)(2,4), so the correct option is A.

Common mistakes

  • Assuming the center can be off the yy-axis. This is wrong because the parabola and the lines y=3xy = \sqrt{3}|x| are symmetric about the yy-axis. Use symmetry first to place the center on the yy-axis.

  • Using the point-to-line distance formula incorrectly for y=3xy = \sqrt{3}x. Write the line as 3xy=0\sqrt{3}x - y = 0 and then apply the distance formula carefully so that the denominator becomes 3+1=2\sqrt{3+1} = 2.

  • Accepting the negative value r=6r = -6 as a radius. Radius must be positive, so reject that case and keep only r=2r = 2.

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