The slope of tangent at any point $(\mathrm{x}, \mathrm{y})$ on a curve $y=y(x)$ is $\frac{x^{2}+y^{2}}{2 x y}, x>0$. If $y(2)=0$, then a value of $y(8)$ is

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Differential Equations

The slope of tangent at any point

(x, y)

on a curve

y = y (x)

\frac{x ^{2} + y ^{2}}{2 x y}, x > 0

. If

y (2) = 0

, then a value of

y (8)

Answer: (B)

Given

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y} = \frac{1 + ( y / x ) ^{2}}{2 ( y / x )}

Let

y = t x

, so

\frac{d y}{d x} = t + x \frac{d t}{d x}

Substituting:

t + x \frac{d t}{d x} = \frac{1 + t ^{2}}{2 t}

x \frac{d t}{d x} = \frac{1 + t ^{2}}{2 t} - t = \frac{1 + t ^{2} - 2 t ^{2}}{2 t} = \frac{1 - t ^{2}}{2 t}

Separating variables:

\int \frac{2 t}{1 - t ^{2}} d t = \int \frac{d x}{x}

- ln ∣1 - t^{2} ∣ = ln ∣ x ∣ + ln ∣ c ∣

ln ∣1 - t^{2} ∣ = - ln ∣ c x ∣

1 - t^{2} = \frac{k}{x}

where k is a constant

1 - \frac{y ^{2}}{x ^{2}} = \frac{k}{x}

Using

y (2) = 0

1 = \frac{k}{2}

, so

k = 2

Thus:

1 - \frac{y ^{2}}{x ^{2}} = \frac{2}{x}

x = 8

1 - \frac{y ^{2}}{64} = \frac{2}{8} = \frac{1}{4}

\frac{y ^{2}}{64} = \frac{3}{4}

y^{2} = 48

y = \pm 43

Since the curve is continuous and

y (2) = 0

, taking the positive value:

y (8) = 43

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option