Answer:
(a)
The equation of the given curve is y=x2−2x+7.
On differentiating with respect to x, we get:
dxdy=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 Slope of the line−1=3−1.
⇒dxdy=2x−2=3−1
