Find the equation of the tangent line to the curve y=x2−2x+7 which is.
(a) parallel to the line 2x−y+9=0.
(b) perpendicular to the line 5y−15x=13.
(a)
The equation of the given curve is y=x2−2x+7.
On differentiating with respect to x, we get:
dydx=2x−2
The equation of the line is 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, at x=2
⇒y=22−2×2+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.
Slope of the line =3
If a tangent is perpendicular to the line 5y−15x=13,
then the slope of the tangent is −1Slope of the line=−13.
⇒dydx=2x−2=−13
⇒2x=−13+2
⇒2x=53
⇒x=56
Now, at x=56
⇒y=2536−106+7=25−60+25236=21736
Thus, the equation of the tangent passing through (56,21736) is given by,
(y−21736)=−13(x−56)
⇒=36y−21736−118(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