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

1 Answer

  • 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}}$
$\Rightarrow y=\large\frac{-x}{3}$$-12$
$\Rightarrow x+3y+36=0$
answered Jun 21, 2013 by sreemathi.v

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