The equation of the tangent to the parabola \(y^2=12 x\), which makes an angle \(30^{\circ}\) with the…
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
\((-9,-6 \sqrt{3})\)
\((9,-6 \sqrt{3})\)
\((-9,6 \sqrt{3})\)
\((9,6 \sqrt{3})\)
Solution
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 the tangent \(x-\sqrt{3} y+9=0\), so on comparing, we get
\(\frac{6}{1}=\frac{-y_1}{-\sqrt{3}}=\frac{6 x_1}{9} \Rightarrow\left(x_1, y_1\right)=(9,6 \sqrt{3})\)
Therefore, point of contact is \((9,6 \sqrt{3})\).