The normal to the curve, `x^(2)+2xy -3y^(2)=0, at (1,1)`

Options

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

B

Explanation

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.