wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Assertion :A normal is drawn at a point P(x,y) of a curve. It meets the x-axis and the y-axis in points A and B, respectively, such that 1OA+1OB=1, where O is the origin. The equation of such a curve passing through (5,4) is (x−1)2+(y−1)2=25 Reason: OA=x+ydydx and OB=(x+ydydx)dydx

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Assertion is incorrect but Reason is correct
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Both Assertion and Reason are incorrect
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
The equation of the normal at (x,y) is,
(Xx)+(Yy)dydx=0

Xx+ydydx+Y(x+ydydx)dydx=1
OA=x+ydydx,OB=x+ydydxdydx
Also, 1OA+1OB=1
1+dydx=x+ydydx(y1)dydx+(x1)=0

Integrating, we get
(y1)2+(x1)2=c
Since the curve passes through (5,4),c=25
Hence, the curve is (x1)2+(y1)2=25
Since the curve passes through (5,4),c=25
Hence, the curve is :
(x1)2+(y1)2=25

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon