Equation of the tangent to the curve `2x^(2)+3y^(2)-5=0` at (1, 1) is

Options

A. `2x-3y-5=0`
B. `2x+3y-5=0`
C. `2x+3y+5=0`
D. `3x+2y+5=0`

Answer

B

Explanation

Equation of the curve is
`2x^(2)+3y^(2)-5=0`
Differentiating w. r.t.x, we get
`4x+6y.(dy)/(dx)=0`
`:.(dy)/(dx)=(-4x)/(6y)=(-2x)/(3y)`
`:.` Slope of the tangent `=(dy)/(dx)=(-2x)/(3y)`
At point `(1,1)((dy)/(dx))_(((1,1)))=(-2)/(3)`
`:.` Equation of tangent at (1,1),
`y-y_(1)=(dy)/(dx)(x-x_(1))`
`y-1=(-2)/(3)(x-1)`
`:.3y-3=-2x+2`
`2x+3y-5=0`