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]}