CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

Tangent to a curve y=f(x) intersects the yaxis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is

A
ydydxx(dydx)2=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
xd2ydx2+(dydx)2=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
ydydx+x=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
xdydx+(dydx)2=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A ydydxx(dydx)2=1
The equation of the tangent at the point
R(x, f(x)) is Yf(x)=f(x)(Xx)
The coordinates of the point P are (0, f(x)xf(x))
The slope of the perpendicular line through P is f(x)xf(x)001=1f(x)
f(x)f(x)x(f(x))2=1
ydydxx(dydx)2=1 which is the required differential equation of the curve y=f(x)

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App