JEE Main 25 Feb 2021 Shift 2 Solved Paper

Curve x2+2y2=2 intersect the line x+y=1 at points P and Q.
First we have to find any common relation between these two curves. Use substitution for the same as follows,
x2+2y2‌‌=2 . . . (i)
x+y‌‌=1,‌ then ‌(x+y)2=12
⇒‌‌x2+y2+2xy=1. . . (ii)
We can write Eq. (i) as,
x2+2y2−2(1)2=0
⇒‌‌x2+2y2−2(x+y)2=0
⇒x2+2y2−2x2−2y2−4xy=0
⇒‌‌−x2−4xy=0⇒−x(x+4y)=0
Gives, x=0 and x+4y=0 or y=‌

−1

4

x
Draw the line y=‌

−1

4

x on graph and take arbitrary point (any one) as follows, From given graph,tan‌θ=‌

1

4

⇒θ=tan−1(‌

1

4

)
We have two lines, y=−‌

1

4

x and x=0 (i.e. Y - axis). Thus, any line joining these two curves makes an angle ‌

π

2

+θ at origin.
∴ Answer is ‌

π

2

+tan−1(‌

1

4

).