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The solution of the equation `(2+x)dy-ydx=0` represents a curve passing through a fixed point P, then the area of equilateral triangle with P as one vertex and `x+y=0` as its one side, isA. `2sqrt3`B. `sqrt3`C. `(2)/(sqrt3)`D. `(4)/(sqrt3)` |
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Answer» Correct Answer - C We have, `(2+x)dy-ydx=0rArr(1)/(y)dx-(1)/(2+x)dx=0` On integrating, we obtain `logy-log(x+2)=logC rArr y=C(x+2)` Clearly, it represents a curve passing through a fixed point `P(-2,0)`. The altitude of the equilateral triangle having one vertex at `P(-2,0)` and opposite side `x+y=0` is `P=|(-2+0)/(sqrt(1+1))|=sqrt2` So, its area is `(p^(2))/(sqrt3)=(2)/(sqrt3)` sq. units |
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