# Find the equation of the tangent to the parabola $y^{2}=16x,$ perpendicular to the line $3x-y+8=0$

Toolbox:
• Equation of any tangent to
• $\quad(i)$ The parabola $y^2=4ax$ is of the form $y=mx+\large\frac{a}{m}$
• $\quad(ii)$ The ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 is of the form y=mx\pm\sqrt{a^2m^2+b^2} • \quad(iii) The hyperbola \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Step 1:
Tangent to $y^2=16x$ perpendicular to $3x-y+8=0$
Compare with $y^2=4ax,4a=16\Rightarrow a=4$
Any tangent to $y^2=4ax$ is $y=mx+\large\frac{a}{m}$
Here $y=mx+\large\frac{4}{m}$
Step 2:
Required tangent is to $3x-y+8=0$ whose slope is $3$.
Therefore slope of the tangent =$\large\frac{-1}{3}$$=m The required tangent is y=-\large\frac{1}{3}$$x+\large\frac{4}{\Large\frac{-1}{3}}$