InterviewSolution
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InterviewSolution
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If a curve passes through the point (1,-2) and has slope of the tangent at any point (x, y) on it as `(x^2_2y)/x`, then the curve also passes through the point -(a) `(sqrt3,0)`(b) `(-1,2) `(c) `(-sqrt2,1)`(d) `(3,0) `A. (a) `(sqrt3,0)`B. (b) (-1,2)C. (c) `(-sqrt2,1)`D. (d) (3,0) |
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Answer» Correct Answer - (a) We know that, slope if the tangent at anypoint (x,y) on the curve is `dy/dx = (x^(2) -2y)/x` (given) `rArr dy/dx + 2/xy=x` shich is a linear differential equation of the form `dy/dx+P(x)cdot y=Q(x),` where ` P(x)=2/xand Q(x) = x` Now, integrating factor `(IF)=e^(int P(x)dx)=e^(int2/xdx)=e^(2log_(e)x` `=e^(2log_(e)x^(2) [therefore m log a =log a^(m)]` `=x^(2) [therefore e^(2log_(e)f(x))=f(x)]` and the solution of differential Eq.(i) is `y(IF) = int Q(x)(IF)dx+ CrArry(x^(2))= int xcdotx^(2)dx+C` `rArr yx^(2) =x^(4)/4+C …(ii)` `therefore` The curve (ii) passes through the point (1,-2), therrefore `-2=1/4+C rArr C= -9/4` `therefore` Equation of required curve is `4yx^(2)+x^(4)-9.` Now, checking all the option, we get only `(sqrt3,0)` satisfy the above equation. |
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