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Q. Let curve $C$ is such that length of subnormal at any point of the curve is 2 and curve passes through the point $P(1,2)$, and $Q(4,4)$ then

Application of Integrals

Solution:

$| my |=2 \Rightarrow ydy =2 dx \Rightarrow \frac{ y ^2}{2}=2 x + C$
$\therefore y ^2=4 x \Rightarrow P \left( t ^2, 2 t \right) $
$\therefore ST =\frac{\frac{2 t }{2}}{\frac{2}{2 t }}=2 t ^2=2$
image
$A=2 \int\limits_0^1 2 \sqrt{x} d x=\frac{4 \cdot\left(x^{\frac{3}{2}}\right)_0^1}{\frac{3}{2}}=\frac{8}{3}$
$\Theta y \frac{d y}{d x}=2$
$\therefore$ Orthogonal trajectory is $y \frac{ dx }{ dy }=-2 \Rightarrow-2 \frac{ dy }{ y }= dx \Rightarrow-2 \ln y = x +\lambda$
$\Rightarrow y =\alpha \cdot e ^{\frac{- x }{2}}$.