\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to…
\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to this line at \(P\) on it, touches the given parabola, then the point \(P\) is
\((-1,-5)\)
\((-1,5)\)
\((-2,-8)\)
\((2,4)\)
Solution
According to the given information, Angle between the tangents is \(90^{\circ}\).
\(\Rightarrow\) The point on the directrix Here,
\(x-2=-3\)
\(x=-1\) is directrix
Given tangent is \(y=3 x-2\)
\(\begin{array}{lrl}
\therefore & y & y(-1)-2=-5 \\
\therefore & \text { Point } P \text { is }(-1,-5).
\end{array}\)