Find the equation of the tangent line to the curve $y = x^{2} - 2 x + 7$ which is perpendicular to the line $5 y - 15 x = 13.$

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Solution

The equation of the line is $5 y - 15 x = 13.$

Slope of the line $= 3$

If a tangent is perpendicular to the line $5 y - 15 x = 13$ ,

then the slope of the tangent is $\frac{- 1}{Slope of the line} = \frac{- 1}{3}$ .

$\Rightarrow \frac{d y}{d x} = 2 x - 2 = \frac{- 1}{3}$

$\Rightarrow 2 x = \frac{- 1}{3} + 2$

$\Rightarrow 2 x = \frac{5}{3}$

$\Rightarrow x = \frac{5}{6}$

Now, at $x = \frac{5}{6}$

$\Rightarrow y = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36}$

Thus, the equation of the tangent passing through $(\frac{5}{6}, \frac{217}{36})$ is given by,

$(y - \frac{217}{36}) = - \frac{1}{3} (x - \frac{5}{6})$

$\Rightarrow= \frac{36 y - 217}{36} - \frac{1}{18} (6 x - 5)$

$\Rightarrow 36 y - 217 = - 2 (6 x - 5)$

$\Rightarrow 36 y - 217 = - 12 x + 10$

$\Rightarrow 36 y + 12 x - 227 = 0$

Hence, the equation of the tangent line to the given curve
(which is perpendicular to line $5 y - 15 x = 13)$ is $36 y + 12 x - 227 = 0$

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Similar questions

Find the equation of the tangent line to the curve y = x² − 2x + 7 which is

(a) parallel to the line 2x − y + 9 = 0

(b) perpendicular to the line 5y − 15x = 13.

Find the equation of the tangent line to the curve $y = x^{2} - 2 x + 7$ which is

(a) parallel to the line 2x-y+9=0.

(a) parallel to the line 5y-15x=13.

y = x^{2} - 2 x + 7

5 y - 15 x = 13

12 x + 36 y - 227 = 0

y = x^{3} + 2 x - 4

x + 14 y + 3 = 0.

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