Find the equation of tangent of the curve `yx^(2)+x^(2)-5x+6=0` at that point at which curve crosses the X-axis.

Explanation

At X-axis, y=0
` :. Yx^(2)+x^(2)-5x+6=0`
`rArr 0+x^(2)-5x+6=0`
`rArr (x-2)(x-3)=0`
`rArr x=2 or x=3`
`:. " Points are "(2,0) or(3,0)`
Again `yx^(2)+x^(2)-5x+6-0`
`rArr x^(2)*(dy)/(dx)+2xy + 2x -5 = 0`
`rArr (dy)/(dx) = (5-2x-2xy)/(x^(2))`
Slope of tangent at point (2,0)
`m= (5-4-0)/4 = 1/4`
and equation of tangent
` y-0= 1/4 (x-2)`
` rArr 4y=x-2 rArr x-4y=2`
Slope of tangent at point (3, 0).
`m=(5-6-0)/3^(2) = - 1/9`
and equation of tangent
` y-0 =-1/9 (x-3)`
` rArr 9y =- x+3 rArr x+9y=3`.