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Q. Given a curve $C$. Suppose that the tangent line at $P(x, y)$ on $C$ is perpendicular to the line joining $P$ and $Q(1,0)$. If the line $2 x+3 y-15=0$ is tangent to the curve $C$ then the curve $C$ denotes.

Differential Equations

Solution:

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