Differential Equations Question 202
Question: Tangent to a curve intercepts the y-axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve
Options:
A) $ y\frac{dy}{dx}-x{{( \frac{dy}{dx} )}^{2}}=1 $
B) $ \frac{xd^{2}y}{dx^{2}}+{{( \frac{dy}{dx} )}^{2}}=0 $
C) $ y\frac{dx}{dy}+x=1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The equation of the tangent at the point $ R(x,f(x)) $ is $ Y-f(x)=f’(x)(X-x) $
The coordinates of the point P are $ (0,f(x)-xf’(x)) $ .
The slope of the perpendicular line through P is $ \frac{f(x)-xf’(x)}{-1}=-\frac{1}{f’(x)} $
Or $ f(x)f’(x)-x{{(f’(x))}^{2}}=1 $ Or $ \frac{ydy}{dx}-x{{( \frac{dy}{dx} )}^{2}}=1 $
Which is the required differential equation to the curve at $ y=f(x). $
BETA
