The equation of the tangent to the parabola \(y^2=16 x\), which is perpendicular to the line \(3 x-4 y+5=0\)…
The equation of the tangent to the parabola \(y^2=16 x\), which is perpendicular to the line \(3 x-4 y+5=0\) is given by
\(4 x-3 y+9=0\)
\(4 x+3 y-9=0\)
\(4 x+3 y+9=0\)
\(4 x-3 y-9=0\)
Solution
Equation of tangent to the parabola \(y^2=16 x\) which is perpendicular to the line \(3 x-4 y+5=0\).
So, slope of tangent is \(\left(-\frac{4}{3}\right)\)
Therefore equation of tangent is
\(\begin{aligned}
& y=-\frac{4}{3} x+\frac{4}{-4 / 3} \Rightarrow y=-\frac{4}{3} x-3 \\
\Rightarrow \quad 4 x+3 y+9 & =0.
\end{aligned}\)