If the line #y=2x+c# touches the ellipse #x^2/4+y^2/3=1#, find the value of #c#?

1 Answer

#c=sqrt19 or c=-sqrt19#

Explanation:

The line #y=2x+c# touches the ellipse , that means there is only one solution of the equations #y=2x+c# and #x^2/4+y^2/3=1#
Now substituting the value of #y# in the equation of ellipse we get
#x^2/4 +(2x+c)^2/3=1#
#=>3x^2+16x^2+4c^2+16cx=12#
#=> 19x^2+16cx+4c^2-12=0#
Now as this equation has only one solution
#:.#Discriminant must be equal to zero
#:.(16c)^2-4×19×(4c^2-12)=0#
#:.256c^2-304c^2+912=0#
#=>48c^2=912#
#:.c=sqrt19 or c=-sqrt19#

graph{(3x^2+4y^2-12)(2x+sqrt19-y)(2x-sqrt19-y)=0 [-5, 5, -2.5, 2.5]}