The equation of the tangent to the curve `y=sqrt(9-2x^(2))` at the point where the ordinate & the abscissa are equal is

Options

A. `2x+y-sqrt3=0`
B. `2x+y-3=0`
C. `2x-y-3sqrt(3)=0`
D. `2x+y-3sqrt(3)=0`

Answer

D

Explanation

Let the point `(x,y)` on tangent to the curve
Given that `x_(1)=y_(1)`
`:.x_(1)=sqrt(9-2x_(1)^(2))`
`impliesx_(1)^(2)=9-2x_(1)^(2)impliesx_(1)= +- sqrt(3)`
Since `ygt0`, therefore the point is `(sqrt(3),sqrt(3))`
Also `y=sqrt(9-2x^(2))impliesy^(2)=9-2x^(2)`
differentiate it
`2y.(dy)/(dx)=-4ximplies(dy)/(dx)=(-2x)/y`
`:.((dy)/(dx))_(((sqrt(3),sqrt(3)))=-2`
So, the equation of tangents is
`(y-sqrt(3))=-2(x-sqrt(3))`
`implies2x+y-3sqrt(3)=0`