JEE Advanced | 2015Straight LineHard

Question

Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y − 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P,Q and R, respectively. Suppose that PQ = PR = If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct expression(s) is(are)

Options

A.e12 + e22 =
B.e1 + e2 =
C.|e12 − e22 | =
D.e1e2 =

Solution

Let E1 :    (a > b)
&  E2 :     (c <  d)
& S : x2 + (y - 1)2 = 2
& tangent to E1, E2 & S is x + y = 3
Now, point of contact of S & tangent is (x1, y1)
Let x = X & y - 1 = Y
∴  X2 + Y2 = 2
& X + Y = 2
Let (X1, Y1) be point of contact.
∴  XX1 + YY1 = 2
∴  X1 = 1 & Y1 = 1
∴  x1 = 1 & y1 = 2
Now, parametric equation of x + y = 3
is 
⇒ 
∴  P ≡ (1, 2),  Q ≡ & R ≡
Now, equation tangent at Q on ellipse E1
        Comparing it with x + y = 3
∴  a2 = 5 & b2 = 4   Now, e12 = 1 -
Similarly, e22 =       ∴ e12e22 =    ⇒   e1e2 =
e12 + e22 = ; |e12 - e22 | =
∴  (A) & (B)

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