15) Find the equation of the tangent line to the curve which is (b) perpendicular to the line

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