JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Ellipse

The line

y = x + 1

meets the ellipse

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

at two points P and Q . If r is the radius of the circle with PQ as diameter then

(3 r)^{2}

is equal to

Answer: (A)

Solution

Substitute

y = x + 1

into the ellipse equation:

\frac{x ^{2}}{4} + \frac{( x + 1 ) ^{2}}{2} = 1

. Multiply by 4:

x^{2} + 2 (x + 1)^{2} = 4

, so

x^{2} + 2 x^{2} + 4 x + 2 = 4

, giving

3 x^{2} + 4 x - 2 = 0

. By Vieta's formulas,

x_{1} + x_{2} = - \frac{4}{3}

and

x_{1} x_{2} = - \frac{2}{3}

. So

(x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4 x_{1} x_{2} = \frac{16}{9} + \frac{8}{3} = \frac{16 + 24}{9} = \frac{40}{9}

. The distance PQ is

∣ x_{1} - x_{2} ∣ 1 + m^{2} = \frac{40}{3} \cdot 2 = \frac{80}{3}

. The diameter is

2 r = \frac{80}{3}

, so

r = \frac{80}{6}

. Then

(3 r)^{2} = (\frac{3 80}{6})^{2} = (\frac{80}{2})^{2} = \frac{80}{4} = 20

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option