TS EAMCET · Maths · Ellipse
The slope of a common tangent to the ellipse \(\frac{x^2}{49}+\frac{y^2}{4}=1\) and the circle \(x^2+y^2=16\) is
- A \(\frac{5}{\sqrt{11}}\)
- B \(\frac{4}{\sqrt{11}}\)
- C \(\frac{3}{\sqrt{11}}\)
- D \(\frac{2}{\sqrt{11}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{\sqrt{11}}\)
Step-by-step Solution
Detailed explanation
Given equation are \[ \begin{aligned} & \frac{x^2}{49}+\frac{y^2}{4}=1 \\ & x^2+y^2=16 \end{aligned} \] and \[ x^2+y^2=16 \] Let equation of common tangent be \[ y=m x+c \] Since, \(y=m x+c\) is a tangent to Eq. (i).…
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