Let A be a point on the x-axis. Common tangents are drawn from A to the curves $\mathrm{x}^{2}+\mathrm{y}^{2}=8$ and $\mathrm{y}^{2}=16x$. If one of these tangents touches the two curves at Q and R, then $(\mathrm{QR})^{2}$ is equal to

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Parabola

Let A be a point on the x-axis. Common tangents are drawn from A to the curves

x^{2} + y^{2} = 8

and

y^{2} = 16 x

. If one of these tangents touches the two curves at Q and R, then

(QR)^{2}

is equal to

(A)64
(B)76
(C)81
(D)72✓ Correct

Answer: (D)

Solution

A tangent to

y^{2} = 16 x

(with

a = 4

) has the form

y = m x + \frac{4}{m}

. For tangency to

x^{2} + y^{2} = 8

, the distance from the origin equals the radius

22

\frac{∣4/ m ∣}{1 + m ^{2}} = 22

, which simplifies to

m^{4} + m^{2} - 2 = 0

, so

m^{2} = 1

. Taking

m = 1

, the tangent

y = x + 4

touches the parabola at

R (4, 8)

and the circle at

Q (- 2, 2)

. Therefore

(QR)^{2} = (4 + 2)^{2} + (8 - 2)^{2} = 36 + 36 = 72

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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