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

The equation of the given curve is y=x2-2x+7
On differentiating with respect to x, we get:
$\frac {dy}{dx}$ = 2x-2

---

(a) The equation of the line is 2x − y + 9 = 0.

2x − y + 9 = 0
y = 2x + 9
This is of the form y = mx + c
∴ Slope of the line = 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:
2 = 2x − 2
2x=4
x=2
Now, x = 2
y = 4 − 4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7) is given by,
y-7=2(x-2)
y-2x-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 .

---

(b) The equation of the line is 5y − 15x = 13

5y-15x=13
y=3x+$\frac {13}{5}$
5y − 15x = 13
∴ This is of the form y = mx + c.
∴Slope of the line = 3 If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent is
$-\frac {1}{slop\ of\ the \ line}=-\frac 13$
2x-2=-$\frac 13$
2x=-$\frac 13$+2
2x=$\frac 53$
x=$\frac 56$
Now, x=$\frac 56$
y=$\frac {25}{36}-\frac {10}{6}+7$= $\frac {25-60+252}{36}$ = $\frac {217}{36}$
Thus, the equation of the tangent passing through($\frac 56$,$\frac {217}{36}$) is given by
y - $\frac {217}{36}$ =-$\frac 13(x-\frac 56)$
$\frac {36y-217}{36 }$= -$\frac {1}{18}$(6x-5)
36y-217 = -2(6x-5)
36y-217 = -12x+10
36y+12x-227 = 0

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