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  2. Find the equation of the tangent line to the curve y = x ^ 2 – 2 x +7 which is perpendicular to the line 5y – 15x = 13.
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15) Find the equation of the tangent line to the curve y = x^2 -2x +7 which is (b) perpendicular to the line 5y - 15x = 13.

Answers (1)
G Gautam harsolia

Perpendicular to line 5y - 15x = 13.\Rightarrow y = 3x + \frac{13}{5}   means slope \ of \ tangent = \frac{-1}{slope \ of \ line}
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
slope \ of \ tangent = \frac{-1}{slope \ of \ line} = \frac{-1}{3}
Now, we know that the  slope of the tangent at a given point to given curve is given by \frac{dy}{dx}
Given the equation of curve is  
y = x^2 - 2x +7
\frac{dy}{dx} = 2x - 2 = \frac{-1}{3}\\ \\ x = \frac{5}{6}
Now, when x = \frac{5}{6} , y = (\frac{5}{6})^2 - 2(\frac{5}{6}) +7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}
Hence, the coordinates are (\frac{5}{6} ,\frac{217}{36})
Now, the equation of tangent passing through (2,7) and with slope m = \frac{-1}{3}  is
y = mx+ c\\ \frac{217}{36}= \frac{-1}{3}\times \frac{5}{6} + c\\ c = \frac{227}{36}
So,
y = \frac{-1}{3}x+\frac{227}{36}\\ 36y + 12x = 227
Hence, equation of tangent is 36y + 12x = 227

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