CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 is


A

ydydxx(dydx)2=1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

yd2ydx2+(dydx)2=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

ydxdy+x=1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

None of these

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(h,f(h)) is yf(h)=f(h)(xh)
The coordinates of the point P are (0,f(h)hf(h))
The slope of the perpendicular line is f(h)+hf(h)
Applying the condition for perpendicularity
f(h)f(h)h(f(h))2=1ydydxx(dydx)2=1
which is the required differential equation to the curve y=f(x)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Formation of Differential Equation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon