The slope of tangent at any point (x, y) on a curve y=y(x) is x^2+y^22xy,x0.

Question

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

Accepted Answer

Let the slope of tangent at any point

$(x, y)$ on a curve $y=y(x)$ is $\frac{d y}{d x}$

According to the question, $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}(x>0)$
[Given]

$$
\begin{aligned}
&\text { Let }  y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
&\Rightarrow  v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\
&\Rightarrow  x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v}
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text { [Taking integration on both sides] } \\
& \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\
& \Rightarrow  \ln \left(1-v^2\right) x=\ln c \quad(c=\text { constant }) \\
& \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\
& \Rightarrow \left(x^2-y^2\right)=c x
\end{aligned}
$$

Put $y(2)=0 \Rightarrow c=2$

$$
\[\begin{gathered} \therefore  x^2-y^2 =2 x \ \therefore  y^2 =x^2-2 x \ \Rightarrow  y(x) = \pm \sqrt{x^2-2 x} \end{gathered}$$
\]

Put $x=8$, we get

$$
\Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3}
$$

Solutions · 1