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Tangent is drawn at any point P of a cur...

Tangent is drawn at any point P of a curve which passes through `(1, 1)` cutting x-axis and y-axis at A and B respectively. If `AP: BP = 3:1`, then,

A

(a) differential equation of the curve is `3x dy/dx+y =0`

B

(b) differential equation of the curve is `3x dy/dx-y =0`

C

(c) curve is passing through `(1/8,2)`

D

(d) normal at `(1,1) is x+3y=4`

Text Solution

Verified by Experts

The correct Answer is:
(a,c)
Since, BP:AP=3:1. Then, equation of tangent is
`Y-y=f'(x)(X-x)`
The intercept on the coordinate axes are
`A(x-y/(f'(x)),0)`
and `B[0,y-xf'(x)]`
Since, P is internally Intercepts a line AB,
`therefore x=(3(x-(y)/(f'(x)))+1xx0)/(3+1)`
`rArr dy/dx=y/(-3x) rArr dy/y=-1/(3x)dx`
On integrating both sides, we get
`therefore xy^(3)= c`
Since, curve passes through (1,1) then c=1.
`therefore xy^(3)=1`
At `x=1/8 rArr y=2`
Hence, (a) and (c) are correct answers.
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