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AP EAMCET · Maths · Ellipse

The equation of a common tangent to the circle \(x^2+y^2=16\) and to the ellipse \(\frac{x^2}{49}+\frac{y^2}{4}=1\) is

  1. A \(y=x+\sqrt{45}\)
  2. B \(y=x+\sqrt{53}\)
  3. C \(\sqrt{11} y=2 x+4\)
  4. D \(\sqrt{11} y=2 x+4 \sqrt{15}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt{11} y=2 x+4 \sqrt{15}\)

Step-by-step Solution

Detailed explanation

\(c^2 = 16(1+m^2)\) \(c^2 = 49m^2+4\) \(16(1+m^2) = 49m^2+4\) \(16+16m^2 = 49m^2+4\) \(33m^2 = 12 \implies m^2 = \frac{4}{11}\) \(m = \pm \frac{2}{\sqrt{11}}\) \(c^2 = 16(1+\frac{4}{11}) = 16(\frac{15}{11})\) \(c = \pm \frac{4\sqrt{15}}{\sqrt{11}}\)…