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.

(a) parallel to the line 5y-15x=13.

Answer 1

The equation of the given curve is $y=x^2-2x+7$

On differentiating w.r.t x, we get $\frac{dy}{dx}=2x-2$

(a) The equation of the line is 2x-y+9=0 $\Rightarrow y=2x+9$

This is of the form y=mx+c

$\therefore$ Slope of the line is m=2

If a tangent is parallel to the line 2x-y+9=0, Then the slope of the tangent is equal to the slope of the line.

Therefore, we have $\frac{dy}{dx}=m\Rightarrow 2x-2=2\Rightarrow 2x=4\Rightarrow x=2$

When x=2 then from Eq (i), we get $y=2^2-2\times 2+7=7$

$\therefore$ The point on the given curve at which tangent is parallel to given line is (2,7) and the equation of the tangent is

y-7=2(x-2) $\Rightarrow 2x-y+3=0$

Hence, the equation of the tangent line to the given curve which is parallel to line 2x-y+9=0 is y-2x+3=0

The equation of the given curve is $y=x^2-2x+7$

On differentiating w.r.t x, we get $\frac{dy}{dx}=2x-2$

(b) The equation of the given line is 5y-15x=13 $\Rightarrow y=3x+\frac{13}{5}$

This is of the form y=mx+c

$\therefore$ Slope of the line is 3.

If a tangent is perpendicular to the line 5y-15x=13, Then the slope of the tangent $=-\frac{1}{3}$

$[\because productofslopes,m_1\times m2=-1\Rightarrow m_2=\frac{-1}{m_1}]$

$\therefore 2x-2=-\frac{1}{3}\Rightarrow 2x=2-\frac{1}{3}=\frac{5}{3}\Rightarrow x=\frac{5}{6}$

When $x=\frac{5}{6}$, then from Eq (i), we get

$y={\left(\frac{5}{6}\right)}^2-2\left(\frac{5}{6}\right)+7=\frac{25}{36}-\frac{5}{7}+7=\frac{25-60+252}{36}=\frac{217}{36}$

$\therefore$ The point on the given curve at which tangent is parallel to given line is $(\frac{5}{6},\frac{217}{36})$ and the equation of the tangent is

$y-\frac{217}{36}=-\frac{1}{3}(x-\frac{5}{6})\Rightarrow y-\frac{217}{36}=-\frac{x}{3}+\frac{5}{18}\Rightarrow 12x+36y-227=0$

Hence, the equation of the tangent line to the given curve which is perpendicular to the line 5y-15x=13 is 36y+12x-227=0.