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The number of values of $\mathrm{c}$ such that the line $y=4 x+c$ touches the curve $\frac{x^{2}}{4}+y^{2}=1$ is
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Verified Answer
The correct answer is:
2
Given, $\mathrm{y}=4 \mathrm{x}+\mathrm{c}$ and $\frac{\mathrm{x}^{2}}{4}+\mathrm{y}^{2}=1$
Condition for tangency,
$$
\begin{array}{ll}
& c^{2}=a^{2} m^{2}+b^{2} \\
\therefore & c^{2}=4(4)^{2}+1^{2} \\
\Rightarrow & c^{2}=65 \\
\Rightarrow & c=\pm \sqrt{65}
\end{array}
$$
Hence, for two values of c, the line touches the curve.
Condition for tangency,
$$
\begin{array}{ll}
& c^{2}=a^{2} m^{2}+b^{2} \\
\therefore & c^{2}=4(4)^{2}+1^{2} \\
\Rightarrow & c^{2}=65 \\
\Rightarrow & c=\pm \sqrt{65}
\end{array}
$$
Hence, for two values of c, the line touches the curve.
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