JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Parabola

Let a common tangent to the curves

y^{2} = 4 x

and

(x - 4)^{2} + y^{2} = 16

touch the curves at the points P and Q. Then

(P Q)^{2}

is equal to ______.

Answer: 32

Solution

For the parabola

y^{2} = 4 x

, a tangent of slope

m

y = m x + \frac{1}{m}

.

For the circle

(x - 4)^{2} + y^{2} = 16

(center

(4, 0)

, radius 4), a tangent of slope

m

y = m (x - 4) \pm 4 1 + m^{2}

.

For a common tangent:

\frac{1}{m} = - 4 m \pm 4 1 + m^{2}

.

Solving (taking the appropriate sign), we get

m = \pm \frac{1}{2 2}

.

For

m = \frac{1}{2 2}

, the point of contact on the parabola is

P = (\frac{1}{m ^{2}}, \frac{2}{m}) = (8, 42)

.

The length of the tangent from P to the circle is

P Q = (8 - 4)^{2} + (42)^{2} - 16 = 16 + 32 - 16 = 32

.

Thus

(P Q)^{2} = 32

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Integer answer

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