1.

The normal to the curve `x^(2) +2xy-3y^(2)=0,` at (1, 1):A. meets the curve again in the third quadrant.B. Meets the curve again the fourth quadrant .C. does not meet the curve again.D. meets the curve again in the second quadrant.

Answer» Correct Answer - B
The equation of the curve is
`x^(2) + 2xy-3y^(2)=0 " " ` …(i)
Differentiating with respect to x, we get
`2x+2y+2x(dy)/(dx)-6y(dy)/(dx)=0rArr (dy)/(dx)=(x+y)/(3y-x) rArr ((dy)/(dx))_((1","1))=1`
The equation of the normal at (1, 1) is
`y-1= -1(x-1) or, (x+y)=2 " " ` ...(ii)
The x-coordinates of the points of intersection of (i) and (ii) are given by
`x^(2)+2x(2-x)-3(2-x)^(2)=0`
`rArr 4x^(2)-16x+12=0`
`rArr x^(2)-4x+3=0 rArr (x-3) (x-1)=0 rArr x=1, 3 `
Now, `x=1 and x+y=2 rArr y=1`
`x=3 and x+y=2 rArr y = -1`
Hence, the normal to the curve meets it again at (3, -1) which lies in the fourth quadrant.


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