53 Find the equation of the curve passing through the point (0,-2) given that at any point (x,y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

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Vishal Baghel | Contributor-Level 10

8 months ago

The slope of the tangent to then curve is $\frac{d y}{d x}$

$\begin{array}{l} \frac{d y}{d x} . y = x \\ \Rightarrow y . d y = x d x \end{array}$

So,

Integrating both sides,

$\begin{array}{l} \int y . d y = \int x d x \\ \Rightarrow \int \frac{y^{2}}{2} = \frac{x^{2}}{2} + c \\ \Rightarrow y^{2} = x^{2} + A, W h e r e, A = 2 c \end{array}$

As the curve passes through (0, -2) we have,

$\begin{array}{l} {(- 2)}^{2} = 0^{2} + A \\ \Rightarrow A = 4 \end{array}$

$∴$ The equation of the curve is

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

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dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.