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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. |
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Answer» 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`. |
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