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The abscissa of a point on the curve `x y=(a+x)^2,`the normal which cuts off numerically equal intercepts from thecoordinate axes, is`-1/(sqrt(2))`(b) `sqrt(2)a`(c) `a/(sqrt(2))`(d) `-sqrt(2)a`A. `(a)/(sqrt(2))`B. aC. `sqrt(2)a `D. `-(a)/(sqrt(2))` |
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Answer» Correct Answer - A::D We have, `xy=(a+x)^(2) " "...(i)` ` rArr y=x+2a+(a^(2))/(x) rArr (dy)/(dx)=1 - (a^(2))/(x^(2)) ` Let `P(x_(1),y_(1)) ` be a point on the curve (i), where the normal cuts off numerically equal intercepts from the coordinate axes. Then, ` (-1)/(((dy)/(dx))_(P))=pm 1 rArr ((dy)/(dx))_(P)=pm 1 rArr 1-(a^(2))/(x_(1)^(2))=pm 1 ` ` rArr 1- (a^(2))/(x_(1)^(2))=1 " or, " 1-(a^(2))/(x_(1)^(2))=-1 ` ` rArr (a^(2))/(x_(1)^(2))=0 " or, " x_(1)=pm (a)/(sqrt(2)) rArr x_(1) = pm (a)/(sqrt(2))` |
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