Question

plss anyone solve ques 23

Answer 1

(1+m^2)+2mcx+(c^2+a^2)=0
Here discriminant=b^2-4ac
D=b^2-4ac
=(2mc)^2-4(1+m^2)(c^2+a^2)
=4m^2c^2-4(1+m^2)(c^2+a^2)
Now put D=0
4m^2c^2-4(1+m^2)(c^2+a^2)=0
4(m^2c^2-(1+m^2)(c^2+a^2))=0
(m^2c^2-(1+m^2)(c^2+a^2))=0
m^2c^2-c^2+a^2-m^2c^2+a^2m^2=0
a^2+a^2m^2=c^2
(1+m^2)a^2=c^2

Hence proved.