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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
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B
1
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C
2
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D
infinite
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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 $${ a }^{ 2 }=4,{ b }^{ 2 }=1,m=4$$
$${ c }^{ 2 }=64+1\Rightarrow c=\pm \sqrt { 65 } $$

$$c=+\sqrt{65} \ \ \ and -\sqrt{65}$$

$$\therefore c \ \ has \ \  2 \ \ values$$

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