Find the equation of the tangent line to the curve y=x2−2x+7 which is parallel to the line 5y-15x=13.
The equation of the given curve is y=x2−2x+7
On differentiating w.r.t x, we get dydx=2x−2
(b) The equation of the given line is 5y-15x=13 ⇒y=3x+135
This is of the form y=mx+c
∴ Slope of the line is 3.
If a tangent is perpendicular to the line 5y-15x=13, Then the slope of the tangent =−13
[∵product of slopes, m1× m2=−1⇒m2=−1m1]
∴2x−2=−13⇒2x=2−13=53⇒x=56
When x=56, then from Eq (i), we get
y=(56)2−2(56)+7=2536−57+7=25−60+25236=21736
∴ The point on the given curve at which tangent is parallel to given line is (56,21736) and the equation of the tangent is
y−21736=−13(x−56)⇒y−21736=−x3+518⇒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.