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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

  1. A \(\frac{5}{\sqrt{11}}\)
  2. B \(\frac{4}{\sqrt{11}}\)
  3. C \(\frac{3}{\sqrt{11}}\)
  4. D \(\frac{2}{\sqrt{11}}\)
Verified Solution

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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