Question

# The number of values of $$c$$ such that the straight line $$y = 4x + c$$ touches the curve $$\dfrac{x^2}{ 4}\,+\, y^2\,=\,1$$ is

A
0
B
1
C
2
D
infinite

Solution

## The correct option is C $$2$$We know that the line $$y=mx+c$$ touches the curve $$\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$ if $${ c }^{ 2 }={ a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 }$$ Here $${ a }^{ 2 }=4,{ b }^{ 2 }=1,m=4$$$${ c }^{ 2 }=64+1\Rightarrow c=\pm \sqrt { 65 }$$$$c=+\sqrt{65} \ \ \ and -\sqrt{65}$$$$\therefore c \ \ has \ \ 2 \ \ values$$Maths

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