Question

Given a curve $C$. Let the tangent line at $P(x,y)$ on $C$ is perpendicular to the line joining $P$ and $Q$ $(1,0)$. If the line $2x+3y-15=0$ is tangent to the curve $C$ then the length of the tangent from the point $(5,0)$ to the curve $C$ is $\sqrt{n}$ (where $n\in N$ ). Find the value of $n$.

Answer 1

Answer: 3

$Y-y=m(X-x)$
now, $\left(\frac{y-0}{x-1}\right)m=-1$
$y\frac{dy}{dx}=1-x$
integrating $\frac{y^2}{2}=x-\frac{x^2}{2}+C$
$x^2+y^2-2x=C$
This is the equation of a circle with centre $(1,0)$ $\therefore 2x+3y-15=0$ is tangent at $(1)$
$\therefore$ perpendicular from $(1,0)$ on the line $=r$
$\left|\frac{2-15}{\sqrt{13}}\right|=\sqrt{1+C}$
$C+1=13\Rightarrow C=12$
hence the curve is $x^2+y^2-2x=12$
Length of tangent is $\sqrt{S_1}=\sqrt{3}\equiv \sqrt{n}$ $\therefore n=3$.