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-3sqrt(3)=0`
B. `2x+y+sqrt(3)=0`
C. `2x+y-sqrt(3)=0`
D. None of these

Answer

A

Explanation

Given `y=sqrt(9-2x^(2))`
According to question ordinate `=` abscissa
`impliesx_(1)=y_(1)`
`:.x_(1)=sqrt(9-2x_(1)^(2))`
`impliesx_(1)^(2)=9=2x_(1)^(2)impliesx_(1)=+-sqrt(3)`
Since `y_(1)gt0`, therefore the point is `(sqrt(3), sqrt(3))`
Now `y=sqrt(9-2x^(2))impliesy^(2)=9-2x^(2)`
On differenting w.r.t `x` we get
`2y (dy)/(dx)=-4x implies (dy)/(dx)=-(2x)/y`
`:.` at `(sqrt(3), sqrt(3)), ((dy)/(dx))_((sqrt(3),sqrt(3))=-2`
So the required equation of tangent is
`(y-sqrt(3))=-2(x-sqrt(3))`
`implies 2x+y-3sqrt(3)=0`