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Let lceiling denote a curve y=y(x) which is in the first quadrant and let the point (1,0) lie on it . Let the tangant ot lceiling at a point P intersect the y-axis at Y_P. If PY_P has length l for each point P on lceiling, then which of the following options is/ are correct ? |
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Answer» `xy'+sqrt(1-x^2)=0` `y-k=((dy)/(dx))_(h,k) (x-h)` ....(i) Now, the tangent (i) intersect the Y-axis at `Y_p` , so coordinates `Y_(p)` is `(0,k-h(dy)/(dx))`, where `(dy)/(dx) =((dy)/(dx))_(h,k)` So, `PY_p=1`(given) `rArr sqrt(h^2+h^2((dy)/(dx))^2)=1` `rArr (dy)/(dx)=+-(sqrt(1-x_1^2))/(x)""["on replacing h by x"]` `rArr dy= +-sqrt(1-x^2)/(x)dx` On PUTTING `x=sintheta, dx =cos thetad theta`,we get `dy =+-sqrt(1-sin^2)/(sin theta)cos theta d theta =+-(cos^2 theta)/(sin theta)d theta` `=+-(cosec theta -sintheta )d theta` `rArr y=+-[In (cosec theta -cot theta )+cos theta]+C` ` y=+-[-In ((1+sqrt((1-x^2)))/(x))+sqrt(1-sin^2)]+C` `rArr y+-[In ((1-costheta)/(sin theta))+cos theta]+C` `rArr y=+-[In (1-sqrt(1-x^2)/(x))+sqrt(1-x^2)]+C ""[because x=sintheta]` ` =+-[-In (1+sqrt(1-x^2))/(x)+sqrt(1-x^2)]+C` [On ratioalization] `because` The curve is in the first QUADRANT so y MUST be positive, so `y=In(1+sqrt((1-x^2))/(x))-sqrt(1-x^2)+C` As curve passes through `(1,0)`, so `0=0-0+c rArr c=0`, so required curve is `y=In (1+sqrt(1-x^2))/(x)=sqrt(1-x^2)` and required differential equation is `(dy)/(dx)=-(sqrt(1-x^2))/(x)` `rArr xy'+sqrt(1-x^2)=0` Hence, options (a) and (c) are CORRECT. |
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