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: $c^2 = (4)(2)^2 + 1$ $c^2 = 4(4) + 1$ $c^2 = 16 + 1 = 17$.

List-$I$	List-$II$
$(i)$ The equation of the major axis	$(p)$ $3x = 34$
$(ii)$ The equation of a directrix	$(q)$ $y = 2$
$(iii)$ The equation of a latus rectum	$(r)$ $x + y = 9$
	$(s)$ $x = 6$
	$(t)$ $x = 3$
	$(u)$ $3y = 34$

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

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