Find the equation of the tangent line to the curve `y=x^2-2x+7` which is (a) parallel to the line `2x -y + 9 = 0` (b) perpendicular to the line `5y -15 x = 13`.

Explanation

Equation of curve
` y=x^(2) - 2x+7` …(1)
` rArr (dy)/(dx) = 2x -2`
(a) Line 2x-y +9= 0
`rArr y = 2x +9`
Its slope m=2
Slope of parallel line = 2
`:. 2x - 2 = 2`
`rArr 2x - 2 = 2`
` rArr 2x = 4 `
` rArr x = 2 `
put x = 2 in equation (1)
` y= 2^(2)-2 xx 2+7=7`
and equation of tangent at point (2, 7)
` y-7=2(x-2)`
` rArr y- 7 = 2x - 4`
`rArr 2x - y + 3 = 0`
(b) Equation of line
`5y - 15x = 13`
` rArr y = 3x + (13)/5`
Its slope = 3
Slope of perpendicular line ` = - 1/3`
`2x - 2 =- 1/3`
` rArr 2x = - 1/3 + 2 = 5/3 `
` rArr x = 5/6`
put ` x= 5/6` in equation (1)
` y=(5/6)^(2)-2xx5/6 + 7 = (25-60+252)/(36) = (217)/36`
and equation of tangent at point `(5/6,(217)/36)`
` y- (217)/36 = - 1/3 (x-5/6)`
(36y- 217)/36 =-((6x-5)/18)`
` rArr 12x+ 36y-227 = 0`