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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. |
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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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