First slide

Formation of differential equations

Question

$A normal at p (x, y) on a curve meets the x -axis at Q and N is the foot of the$ $ordinate at P . If NQ = \frac{x (1 + y^{2})}{1 + x^{2}}, then the equation of the curve given that it passes$ $through the point (3,1) is$

Moderate

A

$x^{2} {-y}^{2} =8$
B

$x^{2} {+2y}^{2} =11$
C

$x^{2} {-5y}^{2} =4$
D

$x^{2} {+5y}^{2} =4$

Solution

$Equation of the normal Y - y = - \frac{dx}{dy} (χ - x)$
$NQ=y \frac{dy}{dx} = \frac{{x(1+y}^{2})}{{1+x}^{2}}$
$\begin{array}{l} \Rightarrow \int \frac{xdx}{1 + x^{2}} = \int \frac{ydy}{1 + y^{2}} \Rightarrow \log (1 + x^{2}) = \log (1 + y^{2}) + \log c \\ \Rightarrow 1 + y^{2} = \frac{1 + x^{2}}{c} pass (3, 1) \Rightarrow c = 5 \\ \Rightarrow 5 + 5 y^{2} = 1 + x^{2} \end{array}$

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