Let x + y = 4 be a normal to a curve y = f (x) at its point (2, 2). If length of the subnormal at any of its point is proportional to the product of abscissa and ordinate of the point and the area enclosed by the curves y = f(x) and y2 – 5y + 4 = 0 is p \sqrt{q} where p and q are coprime numbers then find the value of

Engineering

Mathematics

Variable Separable Differential Equation

Question

Let x + y = 4 be a normal to a curve y = f (x) at its point (2, 2). If length of the subnormal at any of its point is proportional to the product of abscissa and ordinate of the point and the area enclosed by the curves y = f(x) and y² – 5y + 4 = 0 is $p \sqrt{q}$ where p and q are coprime numbers then find the value of | p – q |.

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| m y | = kxy

$1 = 2 k \Rightarrow k = \frac{1}{2}$

$m = \frac{1}{2} x \Rightarrow dy = \frac{x}{2} dx$

$y = \frac{x^{2}}{4} + C$

at (2, 2) ⇒ C = 1

x² = 4 (y – 1)

y² – 5y + 4 = 0 ⇒ y = 1 and y = 4

At y = 4, x² = 12 ⇒ $x = \pm 2 \sqrt{3}$

Required area = $\frac{2}{3}$ × area of the rectangle

$= \frac{2}{3} \times 4 \sqrt{3} \times 3 = 8 \sqrt{3} = p \sqrt{q}$

⇒ | p – q | = 5.