If the X intercept of the tangent to a curve at any point is four times its Y coordinate, then find the equation of the curve.
Let the slope of the tangent at any point $P \left(x,y\right)$ be $= \dfrac{dy}{dx}$
Let the equation of the tangent be $y = x \; \dfrac{dy}{dx} + k$ $\;\;\; \cdots \; (1)$ where k is the intercept.
Given: X intercept of tangent $= 4 y$
$y = 0$ gives the X intercept made by the tangent.
$\therefore$ We have
$0 = 4y \; \dfrac{dy}{dx} + k$
i.e. $k = - 4 \; y \; \dfrac{dy}{dx}$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as
$y = x \; \dfrac{dy}{dx} - 4 \; y \; \dfrac{dy}{dx}$
i.e. $y = \left(x - 4y\right) \dfrac{dy}{dx}$
i.e. $\dfrac{dx}{dy} = \dfrac{x - 4y}{y}$
i.e. $\dfrac{dx}{dy} = \dfrac{x}{y} - 4$ $\;\;\; \cdots \; (3)$
Let $\dfrac{x}{y} = u$ $\;\;\; \cdots \; (4a)$
$\implies$ $x = u \; y$ $\;\;\; \cdots \; (4b)$
Differentiating equation $(4b)$ w.r.t $y$ gives
$\dfrac{dx}{dy} = u + y \; \dfrac{du}{dy}$ $\;\;\; \cdots \; (4c)$
$\therefore$ $\;$ In view of equations $(4a)$ and $(4c)$, equation $(3)$ becomes
$u + y \; \dfrac{du}{dy} = u - 4$
i.e. $y \; \dfrac{du}{dy} = -4$
i.e. $\dfrac{du}{4} = - \dfrac{dy}{y}$ $\;\;\; \cdots \; (5)$
Integrating equation $(5)$ gives
$\dfrac{1}{4} \displaystyle \int du = - \int \dfrac{dy}{y}$
i.e. $\dfrac{u}{4} = - \log \left|y\right| + \log \left|c\right|$ $\;$ $c \;$ is the constant of integration
i.e. $\dfrac{u}{4} = \log \left|\dfrac{c}{y}\right|$
i.e. $\dfrac{x}{4y} = \log \left|\dfrac{c}{y}\right|$ $\;\;\;$ [from equation $(4a)$]
i.e. $e^{\frac{x}{4y}} = \dfrac{c}{y}$
i.e. $y = c \; e^{\frac{-x}{4y}}$ $\;\;\; \cdots \; (6)$
Equation $(6)$ is the required equation of the curve.