Question

If r1 and r2 are distances of points on the ellipse 5x²+5y²+6xy-8=0 which are at maximum and minimum distance from the origin,then r1+r2 is equal to?

Answer 1

The given equation is 5x^2 +6xy +5y^2=8

Let us first rearrange this to make the standard equation of ellipse.

5x^2 + 6xy +5y^2=8
Adding and subtracting 2xy on Left Hand Side,
5x^2 + 5y^2 +8xy - 2xy=8
⇒ x^2 -2xy +y^2 +4x^2 +8xy+4y^2 = 8
⇒ (x-y)^2 +4(x+y)^2 = 8
⇒ [(x-y)^2]/8 +[(x+y)^2]/2 = 1

From this equation, we can see that the major axis of the ellipse is (x+y=0) and the minor axis is (x-y=0)

The two axes intersect at the origin, hence the center of the ellipse is at the origin.

We know that the point on the ellipse closest to the origin will be the point on the ellipse closest to the center, i.e. the point where the minor axis and the ellipse intersect.
Similarly, the point farthest from origin will be the point on the ellipse which also lies on the major axis.

Let the point lying on the major axis be (a,b)
(a,b) also lies on the line x+y=0
⇒ a=-b
The point (-b,b) should satisfy equation of ellipse.
⇒ [(-b-b) ^2]/8 + [(-b+b)^2]/2 =1
⇒ [(-2b)^2]/8 = 1
⇒ 4b^2=8
⇒ b^2=2
⇒ b=+√2 , -√2

Hence the point (a,b) is either (√2, -√2) or (-√2, √2)
Distance of these points from origin= √[(√2)^2 +(-√2)^2]
⇒ r1 = √(2+2)
⇒ r1 = √4
⇒ r1 = 2

Let the point lying on the minor axis be (c,d)
This point lies on the line x=y
⇒ c=d
hence the point becomes (c,c)
This point also satisfies equation of the ellipse.
⇒ [(c-c)^2]/8 + [(c+c)^2]/2 = 1
⇒ 4c^2 = 2
⇒ c^2 = 1/2
⇒ c = 1/√2, or 1/(-√2)

And the point (c,d) is either (1/√2, 1/√2) or (-1/√2, -1/√2)
Distance of point (c,d) from the origin is
r2 = √[(1/√2)^2 +(1/√2)^2]
⇒ r2 = √[(1/2) + (1/2)]
⇒ r2 = √(1)
⇒ r2 = 1

And therefore, r1 +r2 = 2+1 = 3