The equation of the tangent to the parabola \(y^2=12 x\) at \((3,-6)\) is
The equation of the tangent to the parabola \(y^2=12 x\) at \((3,-6)\) is
\(x-y+9=0\)
\(x+y+3=0\)
\(x+y-3=0\)
\(x=3\)
Solution
The equation of tangent to the parabola \(y^2=12 x\) at \((3,-6)\) is
\(\begin{array}{llll}
y(-6) =6(x+3) \\
\Rightarrow x+y+3 =0
\end{array}\)
Hence, option (b) is correct.