Question

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

Answer 1

Consider the line $5x+12y=9$ or $y=\frac{-5}{12}x+\frac{9}{12}$

Hence slope of the line is $\frac{-5}{12}$. Let us assume that the line touches the hyperbola $x^2-9y^2=9$. Therefore it is a tangent to the hyperbola.

Hence at the point of contact the slope $\left(\frac{dy}{dx}\right)$ will be equal to $\frac{-5}{12}$.

Therefore, differentiating the equation of thehyperbola gives us

$2x-18y.y^{\prime }=0$ or $x=9y{y}^{\prime }$ or $y^{\prime }=\frac{x}{9y}=\frac{-5}{12}$ or

$\frac{x}{3y}=\frac{-5}{4}$ or $x=\frac{-15y}{4}$ ...\left(i\right)

Substituting in the equation of the line

$\frac{-75y}{4}+12y=9$ or $(-75+48)y=36$ or $y=\frac{-4}{3}$

Hence $x=5$

Thus the point is $(5,\frac{-4}{3})$.

Substituting in the equation of the hyperbola

$(25-9\left(\frac{16}{9}\right))$

$=25-16$

$=9$

$=RHS$.

Hence the point $(5,\frac{-4}{3})$ is the point of contact of the tangent $5x+12y=9$.

Hence our assumption was indeed correct and hence proved that the lines $5x+12y=9$ touches the hyperbola.