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AP EAMCET · Maths · Parabola

The equation of the tangent to the parabola \(y^2=12 x\), which makes an angle \(30^{\circ}\) with the positive direction of \(X\)-axis is given by \(x-\sqrt{3} y+9=0\), then its points of contact is

  1. A \((-9,-6 \sqrt{3})\)
  2. B \((9,-6 \sqrt{3})\)
  3. C \((-9,6 \sqrt{3})\)
  4. D \((9,6 \sqrt{3})\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((9,6 \sqrt{3})\)

Step-by-step Solution

Detailed explanation

Let the point of tangency of tangent \(x-\sqrt{3} y+9=0\) to the parabola \(y^2=12 x\) is \(\left(x_1 y_1\right)\). Since equation of tangent to the parabola \(y^2=12 x\) at point \(\left(x_1, y_1\right)\) is \(y y_1=6\left(x+x_1\right)\) \(6 x-y_1 y+6 x_1=0\), which represent…