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The normal to the curve `y(x-2)(x-3)=x+6`at the point where the curve intersects the `y-a xi s ,`passes through the point :`(1/2,-1/3)`(2) `(1/2,1/3)`(3) `(-1/2,-1/2)`(4) `((1/(2,1))/2)`A. `(-(1)/(2),-(1)/(2))`B. `((1)/(2),(1)/(2))`C. `((1)/(2),-(1)/(3))`D. `((1)/(2),(1)/(3))` |
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Answer» Correct Answer - B The equation of the curve is `y(x-2)(x-3)=x+6 " " ` …(i) It intersects y-axis at x = 0. Putting x = 0 in (i) , we obtain y = 1. So, the point where (i) cuts y-axis has the coordinates (0, 1). Form (i), we obtain `y=(x+6)/(x^(2)-5x+6)` Differentiating with respect to x, we obtain `(dy)/(dx) = ((x^(2)-5x+6)-(x+6)(2x-5))/((x^(2)-5x+6)^(2))` `therefore ((dy)/(dx))_((0","1))=(6-(-30))/(36) =1` The equation of the normal at (0, 1) is `y-1= -1(x-0) or, x+y-1=0` Clearly, it passes through `((1)/(2), (1)/(2))`. |
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