For what value of c does the line y = 4x + c | vedclass

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(C) 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 $c^2 = a^2m^2 + b^2$.Given the equ

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$2$

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(C) 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 $c^2 = a^2m^2 + b^2$.Given the equation of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$,we have $a^2 = 4$ and $b^2 = 1$.The line equation is $y = 4x + c$,so $m = 4$.Substituting these values into the condition $c^2 = a^2m^2 + b^2$:$c^2 = (4)(4^2) + 1$$c^2 = (4)(16) + 1$$c^2 = 64 + 1 = 65$Thus,$c = \pm \sqrt{65}$.There are $2$ possible values for $c$,which are $\sqrt{65}$ and $-\sqrt{65}$.

For what value of $c$ does the line $y = 4x + c$ touch the curve $\frac{x^2}{4} + y^2 = 1$? Find the number of possible values of $c$.

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