JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
A curve passes through the point $(-1,4)$. If the normal to it at any point $(x,y)$ on it passes through $(2,0)$, then its equation is
Differential Equation
Solution
Concept. The normal at $(x,y)$ has slope $-\frac{1}{y'}$ and passes through $(2,0)$. Formulas used. (1) Slope of the normal $=-\dfrac{1}{dy/dx}$Step 1 - Equation. The slope of the line from $(2,0)$ to $(x,y)$ is $\frac{y}{x-2}$. By (1), $$-\frac{1}{y'}=\frac{y}{x-2}\ \Longrightarrow\ y\,y'=-(x-2).$$ Step 2 - Integrate. $y^2=-(x-2)^2+C$, i.e. $(x-2)^2+y^2=C$. Step 3 - Through $(-1,4)$. $C=9+16=25$. Step 4 - The curve. $(x-2)^2+y^2=25$. Answer: (A)