JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, curves are
A:4x2+9y2=36
B:(2x)2+(2y)2=31
L is common tangent line to curve A and B.
Let slope of L be m.
Given, curve A is equation of ellipse.
‌

x2

9

+‌

y2

4

=1
Then, equation of tangent to ellipse will be
y=mx±√9m2+4. . . (i)
Curve B is circle i.e
x2+y2=‌

31

4

‌ Then, equation of tangent to circle will be ‌
‌‌‌‌y=mx±√(‌

31

4

)m2+(‌

31

4

) . . . (ii)
On comparing Eqs. (i) and (ii), we get On comparing Eqs. (i) and (ii), we get
‌9m2+4=(‌

31

4

)m2+(‌

31

4

)
⇒‌36m2+16=31m2+31⇒5m2=15
⇒‌m2=3
∴ Square of slope of line L is 3 .