TS EAMCET · Maths · Ellipse
The values of c such that the line \(y=4 x+c\) touches the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) is
- A \(\pm 13\)
- B \(\pm 7\)
- C \(\pm \sqrt{65}\)
- D \(\pm \sqrt{74}\)
Answer & Solution
Correct Answer
(C) \(\pm \sqrt{65}\)
Step-by-step Solution
Detailed explanation
\(y=4 x+c\) \(\Rightarrow \quad \frac{x^2}{4}+\frac{y^2}{1}=1\) If \(y=m x+c\) is tangent then \(\begin{aligned} & c^2=a^2 m^2+b^2 \\ & \Rightarrow \quad c^2=4(4)^2+1^2=65 \\ & \therefore \quad c^2= \pm \sqrt{65} .\end{aligned}\)
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