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Q)

Prove that the line $5x+12y=9$ touches the hyperbola $x^{2}-9y^{2}=9$ and find its point of contact. Comment
A)
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 $\big(5,\large\frac{-4}{3}\big)$