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)
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)
(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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