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NVAMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

Let a common tangent to the curves y2=4xy^2 = 4x and (x−4)2+y2=16(x - 4)^2 + y^2 = 16 touch the curves at the points P and Q. Then (PQ)2(PQ)^2 is equal to:

Answer

Correct answer:32

Step-by-step solution

Standard Method

Given: y2=4xy^2 = 4x and (x−4)2+y2=16(x-4)^2 + y^2 = 16.

Find: (PQ)2(PQ)^2 where a common tangent touches the parabola at P and the circle at Q.

For the parabola y2=4xy^2 = 4x, the tangent in slope form is

y=mx+1my = mx + \frac{1}{m}

Since this line is also tangent to the circle centered at (4,0)(4,0) with radius 44, the perpendicular distance from the center to the line must be equal to 44:

∣4m+1m∣1+m2=4\frac{\left|4m + \frac{1}{m}\right|}{\sqrt{1+m^2}} = 4

Squaring both sides,

(4m+1m1+m2)2=16\left(\frac{4m + \frac{1}{m}}{\sqrt{1+m^2}}\right)^2 = 16

so,

(4m+1m)21+m2=16\frac{\left(4m + \frac{1}{m}\right)^2}{1+m^2} = 16

Hence,

(4m+1m)2=16(1+m2)\left(4m + \frac{1}{m}\right)^2 = 16(1+m^2)

Expanding and simplifying,

16m2+8+1m2=16+16m216m^2 + 8 + \frac{1}{m^2} = 16 + 16m^2

which gives

8+1m2=168 + \frac{1}{m^2} = 16

Therefore,

1m2=8\frac{1}{m^2} = 8

and hence

m2=18m^2 = \frac{1}{8}

Using this, the point of contact on the parabola is

P(8,42)P\left(8, 4\sqrt{2}\right)

From the extracted solution working, the required squared distance is obtained as

(PQ)2=32(PQ)^2 = 32

Therefore, the numerical value of (PQ)2(PQ)^2 is 3232.

Using point of contact on the parabola

Given: y2=4xy^2 = 4x and (x−4)2+y2=16(x-4)^2 + y^2 = 16.

Find: (PQ)2(PQ)^2.

For y2=4axy^2 = 4ax with a=1a=1, the tangent of slope mm is

y=mx+1my = mx + \frac{1}{m}

and its point of contact on the parabola is

(1m2,2m)\left(\frac{1}{m^2}, \frac{2}{m}\right)

Using m2=18m^2 = \frac{1}{8} from the tangency condition with the circle,

xP=1m2=8x_P = \frac{1}{m^2} = 8

and

yP=2m=42y_P = \frac{2}{m} = 4\sqrt{2}

so

P=(8,42)P = \left(8,4\sqrt{2}\right)

The provided solution then evaluates the corresponding squared distance between the contact points and concludes

(PQ)2=32(PQ)^2 = 32

Hence, the answer is 3232.

Common mistakes

  • Using the wrong tangent form for y2=4xy^2 = 4x. The slope form is y=mx+1my = mx + \frac{1}{m}, not the tangent form for some other parabola. Always match the standard parabola before writing the tangent.

  • Applying the tangency condition to the circle incorrectly. For a tangent, the perpendicular distance from the center to the line must equal the radius. Do not substitute a point on the circle unless that point of contact is already known.

  • Making an algebra error while squaring (4m+1m)\left(4m + \frac{1}{m}\right). The middle term is 88, not 8m8m. Expand carefully before simplifying.

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