Question

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 $5y-15x=13$

Answer 1

$a.$
Given curve: $y=x^2-2x+7$
Now, $\frac{dy}{dx}=2x-2$
Line: $2x-y+9=0$
$\Rightarrow y=2x-9$
Slope of this line is $2$
Tangent is parallel to this line, so slopes of both the lines are same, $\frac{dy}{dx}=2$
$\Rightarrow 2x-2=2$
$\Rightarrow x=2$
From the equation of curve $y=x^2-2x+7$

$\Rightarrow y=2^2-2\left(2\right)+7$
$\Rightarrow y=7$
Thus, the point is $(2,7)$.
Equation of tangent through $(2,7)\&$ having slope is $2$ is $(y-7)=2(x-2)$
$\Rightarrow y-7=2x-4$
$\Rightarrow y-2x-3=0$
Hence, the required equation of tangent is $y-2x-3=0$

$b.$
Given curve:
$y=x^2-2x+7$
Now, $\frac{dy}{dx}=2x-2$
Line: $5y-15x=13$
$\Rightarrow y=3x+\frac{13}{5}$
Slope of this line is $3$.
Tangent is perpendicular to this line, so
Slope of tangent $=-\frac{1}{3}$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{3}$
$\Rightarrow 2x-2=-\frac{1}{3}$
$\Rightarrow x=\frac{5}{6}$
From the equation of curve $y=x^2-2x+7$
$y={\left(\frac{5}{6}\right)}^2-2\left(\frac{5}{6}\right)+7$
$\Rightarrow y=\frac{25}{36}-\frac{5}{3}+7$
$\Rightarrow y=\frac{217}{36}$
Thus, the point is $(\frac{5}{6},\frac{217}{36})$
Equation of tangent through $(\frac{5}{6},\frac{217}{36})\&$ having slope is $-\frac{1}{3}$ is
$(y-\frac{217}{36})=-\frac{1}{3}(x-\frac{5}{6})$
$\Rightarrow \frac{36y-217}{36}=\frac{-1}{3}(x-\frac{5}{6})$
$\Rightarrow 36y-217=-12(x-\frac{5}{6})$
$\Rightarrow 36y-217=-12x+10$
$\Rightarrow 36y+12x-227=0$
Hence, the required equation of tangent is $36y+12x-227=0$.