KCET · Maths · Parabola
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
- A 1
- B 2
- C \(\infty\)
- D 0
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
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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