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

0

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1

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2

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infinite

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Solution

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

$c = + \sqrt{65} a n d - \sqrt{65}$

$∴ c h a s 2 v a l u e s$

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Similar questions

c

y = 2 x + c

4 x^{2} + y^{2} = 4

C

y = 4 x + c

\frac{x^{2}}{4} + y^{2} = 1

3 x + 4 y = c

\frac{x^{4}}{2} = x + y

y = 4 x + m

x^{2} + 4 y^{2} = 4

x^{2} + y^{2} - 4 y = 0,

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