It is given that x+y+n=0 is normal to the ellipse x2+3y2=3 this implies, n=±
(3−1)
√3+1
=±1 since n cannot be negative then, n=1 Consider the equation of normal, x+y+1=0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) The equation x+my+3=0 is tangent of ellipse x2+5y2=5 so,
3
m
=±√
5
m2
+1
9
m2
=
5
m2
+1 m2=4 m=±2 The m cannot be greater than zerothen, m=−2 Consider equation of tangent, x−2y+3=0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) Solve equation (I) and (II), x=−
5
3
,y=
2
3
The above values satisfy the equation, x−5y+5=0 Therefore, the correct option is (2)