The number of values of c such that the line | vedclass

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(B) The equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$,where $a^{2}=4$ and $b^{2}=1$.The equation of the line is $y=mx+c$,where $m=4$.The condit

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$2$

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(B) The equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$,where $a^{2}=4$ and $b^{2}=1$.The equation of the line is $y=mx+c$,where $m=4$.The condition for the line $y=mx+c$ to be tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.Substituting the values,we get $c^{2}=4(4)^{2}+1$.$c^{2}=4(16)+1=64+1=65$.Therefore,$c=\pm \sqrt{65}$.Since there are two possible values for $c$,the line touches the curve for $2$ values of $c$.

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

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