JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Ellipse

A common tangent

T

to the curves

C_{1} : \frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

and

C_{2} : \frac{x ^{2}}{42} - \frac{y ^{2}}{143} = 1

does not pass through the fourth quadrant. If

T

touches

C_{1}

(x_{1}, y_{1})

and

C_{2}

(x_{2}, y_{2})

, then

∣2 x_{1} + x_{2} ∣

is equal to ____.

Answer: 20

Solution

Common tangent to ellipse

C_{1}

y = m x \pm 4 m^{2} + 9

.

Common tangent to hyperbola

C_{2}

y = m x \pm 42 m^{2} - 143

.

Equating:

4 m^{2} + 9 = 42 m^{2} - 143 \Rightarrow 38 m^{2} = 152 \Rightarrow m = \pm 2

.

For

m = 2

c = \pm 5

. Since tangent doesn't pass through fourth quadrant:

T : y = 2 x + 5

.

For

C_{1}

: comparing with

\frac{x x _{1}}{4} + \frac{y y _{1}}{9} = 1

, we get

x_{1} = - \frac{8}{5}

.

For

C_{2}

: comparing with

\frac{x x _{2}}{42} - \frac{y y _{2}}{143} = 1

, we get

x_{2} = - \frac{84}{5}

.

Therefore:

∣2 x_{1} + x_{2} ∣ = - \frac{16}{5} - \frac{84}{5} = - \frac{100}{5} = 20

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer