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Q.
The line normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4) then its equation is
Differential Equations
Solution:
The equation of normal to a curve at a point (x, y) is
$\left(Y-y\right) \frac{dy}{dx} + \left(X-x\right) = 0$
Since it passes throgh the point $\left(3, 0\right)$, we have
$\left(0-y\right) \frac{dy}{dx} + \left(3-x\right) = 0 \Rightarrow y \frac {dy}{dx} = \left(3-x\right) \Rightarrow ydy = \left(3-x\right) dx$
Integrating, we get $\frac{y^{2}}{2} = 3x-\frac{x^{2}}{2} +C$
$\Rightarrow x^{2} + y^{2} - 6x - 2c = 0$
Since the curve passes through $\left(3, 4\right)$, we have
$9 + 16 - 18 - 2c = 0 \Rightarrow c = \frac{7}{2}$.
$\therefore x^{2 }+ y^{2} - 6x - 7 = 0$ is the required equation of the curve.