If $\mathrm{y}=\mathrm{m}_{1} \mathrm{x}+\mathrm{c}_{1}$ and $\mathrm{y}=\mathrm{m}_{2} \mathrm{x}+\mathrm{c}_{2}, \mathrm{~m}_{1} \neq \mathrm{m}_{2}$ are two common tangents of circle $x^{2}+y^{2}=2$ and parabola $y^{2}=x$, then the value of $8\left|m_{1} m_{2}\right|$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Parabola

y = m_{1} x + c_{1}

and

y = m_{2} x + c_{2}, m_{1} \neq = m_{2}

are two common tangents of circle

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

and parabola

y^{2} = x

, then the value of

8 ∣ m_{1} m_{2} ∣

is equal to

Answer: (C)

The tangent to parabola

y^{2} = x

y = m x + \frac{1}{4 m}

.

For this to be tangent to circle

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

, the distance from center

(0, 0)

to the line must equal

2

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

Squaring:

\frac{1}{16 m ^{2} ( 1 + m ^{2} )} = 2

1 = 32 m^{2} + 32 m^{4}

32 m^{4} + 32 m^{2} - 1 = 0

Using quadratic formula in

m^{2}

m^{2} = \frac{- 32 \pm 1024 + 128}{64} = \frac{- 32 \pm 1152}{64} = \frac{- 32 \pm 24 2}{64} = \frac{- 4 \pm 3 2}{8}

Taking the positive root:

m^{2} = \frac{3 2 - 4}{8}

Since there are two tangents with slopes

m_{1}

and

m_{2}

m_{1} m_{2} = - \frac{3 2 - 4}{8}

(considering both signs)

Therefore

8∣ m_{1} m_{2} ∣ = 32 - 4 = - 4 + 32

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option