Application of Derivatives 1 Question 10

10. The normal to the curve x2+2xy3y2=0 at (1,1)

(a) does not meet the curve again

(2015 Main)

(b) meets in the curve again the second quadrant

(c) meets the curve again in the third quadrant

(d) meets the curve again in the fourth quadrant

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Answer:

Correct Answer: 10. a=12,b=34,c=3

Solution:

  1. Given equation of curve is

x2+2xy3y2=0

On differentiating w.r.t x, we get

2x+2xy+2y6yy=0y=x+y3yx At x=1,y=1,y=1 i.e. dydx(1,1)=1

Equation of normal at (1,1) is

y1=11(x1)y1=(x1)x+y=2

On solving Eqs. (i) and (ii) simultaneously, we get

x2+2x(2x)33(2x)2=0x2+4x2x23(4+x24x)=0x2+4x123x2+12x=04x2+16x12=04x216x+12=0x24x+3=0(11)(x3)=0x=1,3

Now, when x=1, then y=1

and when x=3, then y=1

P=(1,1) and Q=(3,1)

Hence, normal meets the curve again at (3,1) in fourth quadrant.

Alternate Solution

 Given, x2+2xy3y2=0(xy)(x+3y)=0xy=0 or x+3y=0

Equation of normal at (1,1) is

y1=1(x1)x+y2=0

It intersects x+3y=0 at (3,1) and hence normal meets the curve in fourth quadrant.



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