Question

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

• A. 20
• B. 12
• C. 11
• D. 8

Answer 1

Answer: (A)

20

Answer 2

The correct option is(A): 20.

Let point (a , a + 1) as the point of intersection of line and ellipse.

So,

$\frac{a^2}{4}+\frac{a+1}{2}=1⇒a^2+2(a^2+2a+1)=4$

$⇒3a^2+4a-2=0$

If the roots of this equation are α and β.

So, P(α, α + 1) and Q(β, β + 1)

PQ 2 = 4 r 2 = (α – β)2 + (α – β)2

$⇒9r^2=\frac{9}{4}(2(α-β)^2)$

$=\frac{9}{2}[(α+β)^2-4αβ]$

$=\frac{9}{2}[(-\frac{4}{3})^2+\frac{8}{3}]$

$=\frac{1}{2}[16+24]=20$