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MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

The shortest distance between the curves y2=8xy^2 = 8x and x2+y2+12y+35=0x^2 + y^2 + 12y + 35 = 0 is:

  • A

    23−12\sqrt{3} - 1

  • B

    2\sqrt{2}

  • C

    32−13\sqrt{2} - 1

  • D

    22−12\sqrt{2} - 1

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The curves are y2=8xy^2 = 8x and x2+y2+12y+35=0x^2 + y^2 + 12y + 35 = 0.

Find: The shortest distance between them.

First rewrite the second curve as a circle by completing the square:

x2+y2+12y+35=0x^2 + y^2 + 12y + 35 = 0 x2+y2+12y=−35x^2 + y^2 + 12y = -35 x2+(y+6)2−36=−35x^2 + (y+6)^2 - 36 = -35 x2+(y+6)2=1x^2 + (y+6)^2 = 1

So the circle has centre C(0,−6)C(0,-6) and radius r=1r = 1.

For the parabola y2=8xy^2 = 8x, comparing with y2=4axy^2 = 4ax gives a=2a = 2.

Take the parametric point on the parabola as P(am2,−2am)P(am^2,-2am). With a=2a=2, this becomes

P(2m2,−4m)P(2m^2,-4m)

The normal at this point is

y=mx−2am−am3y = mx - 2am - am^3

so here

y=mx−4m−2m3y = mx - 4m - 2m^3

Normal Through Circle Centre

For the shortest distance from a point on the parabola to the circle, the joining line must pass through the centre of the circle. Hence the normal to the parabola at the nearest point passes through C(0,−6)C(0,-6).

Substitute C(0,−6)C(0,-6) into the normal:

−6=m(0)−4m−2m3-6 = m(0) - 4m - 2m^3 2m3+4m−6=02m^3 + 4m - 6 = 0 m3+2m−3=0m^3 + 2m - 3 = 0

By inspection, m=1m=1 is a root. The remaining factor has no real root, so the real value is m=1m=1.

Therefore the point on the parabola is

P(2(1)2,−4(1))=(2,−4)P(2(1)^2,-4(1)) = (2,-4)

Distance from Centre Minus Radius

Now compute the distance from P(2,−4)P(2,-4) to the centre C(0,−6)C(0,-6):

PC=(2−0)2+(−4−(−6))2PC = \sqrt{(2-0)^2 + \left(-4-(-6)\right)^2} PC=4+4=8=22PC = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}

Since the circle has radius 11, the shortest distance between the parabola and the circle is

PC−r=22−1PC - r = 2\sqrt{2} - 1

Therefore, the correct option is D.

The first approach shown in the source concludes incorrectly with intermediate reasoning near the origin, but the detailed normal-based working and the stated correct option both give the consistent result 22−12\sqrt{2} - 1.

Common mistakes

  • Choosing the vertex (0,0)(0,0) of the parabola as the nearest point to the circle is incorrect because the shortest distance between two curves need not occur at the nearest-looking extremity. The correct approach is to use the normal condition at the nearest point.

  • Not converting x2+y2+12y+35=0x^2 + y^2 + 12y + 35 = 0 into standard circle form leads to a wrong centre and radius. Complete the square to get x2+(y+6)2=1x^2 + (y+6)^2 = 1, so the centre is (0,−6)(0,-6) and radius is 11.

  • Using the tangent instead of the normal for the parabola is wrong. At the point of minimum distance, the joining segment to the other curve is along the normal direction.

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