Explanation:
C Given, parabola, \(\mathrm{y}^2=16 \mathrm{x}=4(4) \mathrm{x}\)
line: \(3 x-4 y+5=0\)
\(y=\frac{3}{4} x+\frac{5}{4}\)
slope, \(\mathrm{m}=\frac{3}{4}\), slope of tangent \(=-\frac{4}{3}\)
Now, differentiate parabola w.r.t. \(\mathrm{x}\)
\(2 \mathrm{y}_1 \frac{\mathrm{dy}}{\mathrm{dx}}=16\)
\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8}{\mathrm{y}_1}=\frac{-4}{3} \text { (slope) }\)
\(\Rightarrow \mathrm{y}_1=\frac{-2}{3}\)
\(\therefore \quad \mathrm{x}_1=\frac{\mathrm{y}_1^2}{16}=\frac{\left(\frac{-2}{3}\right)^2}{16}\)
\(\mathrm{x}_1=\frac{1}{36}\)
\(\therefore\) Equation of tangent from point \(\left(\frac{1}{36}, \frac{-2}{3}\right)\) is,
\(\mathrm{yy}_1=8\left(\mathrm{x}+\mathrm{x}_1\right)\)
\(\left(\frac{-2}{3}\right) \mathrm{y}=8\left(\mathrm{x}+\frac{1}{36}\right)\)
\(\Rightarrow \quad 8 \mathrm{x}+\frac{2}{3} \mathrm{y}+\frac{2}{9}=0\)
\(\Rightarrow \quad 4 \mathrm{x}+3 \mathrm{y}+9=0\)