The abscissa of a point on the curve `x y=(a+x)^2,` the normal which cuts off numerically equal intercepts from the coordinate axes, is (a) `-1/(sqrt(2))` (b) `sqrt(2)a` (c) `a/(sqrt(2))` (d) `-sqrt(2)a`

Options

A. `(a)/(sqrt(2))`
B. a
C. `sqrt(2)a `
D. `-(a)/(sqrt(2))`

Answer

A, D

Explanation

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))`