If the line y=2x+c touches the curve x^2+4y^2 | vedclass

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(B) The given equation of the curve is $x^2+4y^2=4$. Dividing by $4$,we get the standard form of the ellipse: $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here,$a

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$17$

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(B) The given equation of the curve is $x^2+4y^2=4$. Dividing by $4$,we get the standard form of the ellipse: $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here,$a^2=4$ and $b^2=1$. The condition for the line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$. Comparing $y=2x+c$ with $y=mx+c$,we have $m=2$. Substituting the values $a^2=4$,$b^2=1$,and $m=2$ into the condition: $c^2 = (4)(2)^2 + 1$ $c^2 = 4(4) + 1$ $c^2 = 16 + 1 = 17$.

If the line $y=2x+c$ touches the curve $x^2+4y^2=4$,then $c^2=$

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