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Find the equation of the tangents to the parabola $y^{2}=6x,$ parallel to $3x-2y+5=0$

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  • 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 $y^2=6x$ parallel to $3x-2y+5=0$
Compare with $y^2=4ax$ we have $4a=6$ or $a=\large\frac{3}{2}$
Any tangent to $y^2=4ax$ to $y=mx+\large\frac{a}{m}$
Here $y=mx+\large\frac{3}{2m}$
Step 2:
The required tangent is parallel to $3x-2y+5=0$ whose slope is $\large\frac{3}{2}$.
Therefore $m=\large\frac{3}{2}$
The required tangent is $y=\large\frac{3}{2}$$x+\large\frac{3}{2\Large\frac{3}{2}}$
$y=\large\frac{3}{2}$$x+1$
$2y=3x+2$
$\Rightarrow 3x-2y+2=0$
answered Jun 21, 2013 by sreemathi.v
 

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