On differentiating with respect to x,we get:
(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
Now, x = 2
y = 4 −4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7)is given by,
Hence, the equation of thetangent line to the given curve (which is parallel to line 2x − y +9 = 0) is
.
(b) The equation of the line is 5y −15x = 13.
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
Thus, the equation of thetangent passing through
is given by,
Hence, the equation of thetangent line to the given curve (which is perpendicular to line 5y −15x = 13) is
