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Find the equation of the tangent and normal to the hyperbola $4x^{2}-y^{2}=64 $ ,which are parallel to $10x-3y+9=0$

1 Answer

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:
$4x^2-y^2=64$
The above equation is divided by $64$
$\large\frac{x^2}{16}-\frac{y^2}{64}$$=1$
$a^2=16$,$b^2=64$
Step 2:
The tangent to $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Here $a^2=16$,$b^2=64$
Any tangent to the given hyperbola is of the form $y=mx\pm\sqrt{16m^2-64}$
Step 3:
The required tangents are parallel to $10x-3y+9=0$ whose slope is $\large\frac{10}{3}$$=m$
The equations of the tangents are $y=\large\frac{10x}{3}$$\pm\sqrt{16\times \large\frac{100}{9}-\normalsize 64}$
$y=\large\frac{10x}{3}$$\pm\sqrt{ \large\frac{1600-576}{3} }$
$3y=10x\pm \sqrt{1024}$
$3y=10x\pm32$
answered Jun 21, 2013 by sreemathi.v
 

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