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Prove that the line $5x+12y=9$ touches the hyperbola $x^{2}-9y^{2}=9$ and find its point of contact.

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Toolbox:
  • For $y=mx+c$ to be a tangent to
  • $\quad(i) \;y^2=4ax,c=\large\frac{a}{m}$ and the point of contact is $\big(\large\frac{a}{m^2},\large\frac{2a}{m}\big)$
  • $\quad(ii)\;\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1,c^2=a^2m^2+b^2$ and the point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{b^2}{c}\big)$
  • $\quad(iii)\;\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1,c^2=a^2m^2-b^2$ and the point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{-b^2}{c}\big)$
Step 1:
The line $y=mx+c$ touches the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ if $c^2=a^2m^2-b^2$
Step 2:
Here the equation of the line is $5x+12y=9$
$y=\large\frac{-5x}{12}+\frac{9}{12}$
$m=\large\frac{-5}{12}$ and $c=\large\frac{9}{12}=\large\frac{3}{4}$
$c^2=\large\frac{9}{12}$
Step 3:
The hyperbola is $x^2-9y^2=9$
This equation is divided by $9$
$\large\frac{x^2}{9}$$-y^2=1$
$a^2=9$ and $b^2=1$
$a^2m^2+b^2=9\times \large\frac{25}{144}$$-1$
$\Rightarrow \large\frac{25-16}{16}=\large\frac{9}{16}$$=c^2$
Step 4:
$c^2=a^2m^2+b^2\Rightarrow $ the line is a tangent to the hyperbola.
The point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{-b^2}{c}\big)$
$\large\frac{-a^2m}{c}=\large\frac{-9\times \Large\frac{-5}{12}}{\Large\frac{3}{4}}$$\Rightarrow 5$
$\large\frac{-b^2}{c}=\frac{-1}{\Large\frac{3}{4}}=\frac{-4}{3}$
The point of contact is $\big(5,\large\frac{-4}{3}\big)$
answered Jun 21, 2013 by sreemathi.v
 

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