The correct option is D Centre: (0,0)
Given: x2+2xy+2y2−1=0
To Find: (i) Centre of the ellipse (ii) Length of major and minor axes
Step-1: Recall the standard equation of conic
Step-2: Recall the method od partial differentiation and find the centre
Step-3: Assume parametric coordinates of any point on ellipse and find major and minor axes
(i) Centre of the ellipse
The general 2nd degree equation of a conic is ax2+2hxy+by2+2gx+2fy+c=0
The centre of this equation can be found using partial differentiation.
δf(x,y)δx=2x+2y=0
δf(x,y)δy=2x+4y=0
On solving the two equations we get x=0 and y=0
So, the centre is (0,0).
(ii) Length of major and minor axes
Let P(rcosθ,rsinθ) be any point on the given ellipse.
r= Maximum distance of point P from the centre (0,0)
(rcosθ)2+2(rsinθ)2+2rcosθ rsinθ−1=0
⇒r2cos2θ+2r2sin2θ+2r2cosθsinθ−1=0
⇒r2(cos2θ+sin2θ)+r2sin2θ+2r2cosθ sinθ−1=0
⇒r2+r2(1−cos2θ2)+2r2sin2θ−2=0
⇒3r2+r2(2sin2θ−cos2θ)=2
⇒r2=23+2sin2θ−cos2θ
Maximum length of r = Length of semi - major axis
Minimum length of r = Length of semi - minor axis
−√22+(−1)2≤2sin2θ−cos2θ≤√22+(−1)2
−√5≤2sin2θ−cos2θ≤√5
r2max=23−√5 and r2min=23+√5
rmax=√2√3−√5 and rmin=√2√3+√5
Length of major - axis =2rmax=2√2√3−√5
Length of minor - axis =2rmin=2√2√3+√5