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AP EAMCET · Maths · Ellipse

The equation of the tangent to the ellipse \(x^2+16 y^2=16\) which makes an angle \(60^{\circ}\) with the \(X\)-axis is

  1. A \(\sqrt{3} x-y+7=0\)
  2. B ) \(\sqrt{3} x+y+7=0\)
  3. C \(\sqrt{3} x+y-7=0\)
  4. D \(\sqrt{3} x-y=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{3} x-y+7=0\)

Step-by-step Solution

Detailed explanation

Equation of tangent to the ellipse \(\frac{x^2}{16}+\frac{y^2}{1}=1\), have slope \(m=\tan 60^{\circ}=\sqrt{3}\) is \(\begin{aligned} & y=\sqrt{3} x \pm \sqrt{48+1} \\ \Rightarrow \quad & \sqrt{3} x-y+7=0 \text { or } \sqrt{3} x-y-7=0\end{aligned}\)