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 $5y - 15x = 13$.

We have,
$y = {x^2} - 2x + 7$ …(i)
Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 2x - 2$

(a) The slope of the tangent to the curve (i) is $2x - 2$

Slope of line $2x + y - 9 = 0$ is $2$

Since the tangent is parallel to the line $2x + y - 9 = 0$,

Therefore their slopes are equal $\Rightarrow 2x - 2 = 2 \Rightarrow x = 2$
Putting $x = 2$in (i),we get $y = 7$.

Therefore the equation of tangent at $\left( {2,7} \right)$ parallel to $2x + y - 9 = 0$is
$\left( {y - 7} \right) = 2\left( {x - 2} \right) \Rightarrow y - 7 = 2x - 4 \Rightarrow 2x - y + 3 = 0$

(b) S1ope of tangent to curve (i) is $2x - 2$
Slope of line $5y - 15x = 13$ is $3$

Since the required tangent is perpendicular to the line $5y - 15x = 13$
Therefore the product of their slopes is $- 1$
$\Rightarrow (2x - 2)(3) = - 1 \Rightarrow 6x - 6 = - 1 \Rightarrow 6x = 5 \Rightarrow x = \cfrac{5}{6}$

Putting $x = \cfrac{5}{6}$ in (i), we get $y = \cfrac{{217}}{{36}}$

Also, slope of the required tangent $= \cfrac{{ - 1}}{3}$

Therefore the equation of tangent at $(\cfrac{5}{6},\cfrac{{217}}{{36}})$ perpendicular to $5y - 15x = 13$ is
$\left( {y - \cfrac{{217}}{{36}}} \right) = - \cfrac{1}{3}\left( {x - \cfrac{5}{6}} \right) \Rightarrow 12x + 36y - 227 = 0$