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)
& E2 :
& 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 ≡
Now, equation tangent at Q on ellipse E1
∴ a2 = 5 & b2 = 4 Now, e12 = 1 -
Similarly, e22 =
e12 + e22 =
∴ (A) & (B)
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