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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of the tangent line to the curve \(y = x^2 - 2x +7\) which is $(b)\; perpendicular\; to\; the\; line\; 5y-15x=13$

This is (b) part of the multi-part question q15

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1 Answer

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given :$y=x^2-2x+7$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=$$(2x-2)$
$\quad\;\;=2(x-1)$
Step 2:
When the tangent is perpendicular to the line $5y-15x=13$
Since they are perpendicular,
slope of the tangent $\times$ slope of the line =$-1$
Slope of the line is $-\big(\large\frac{-15}{5}\big)$
$\Rightarrow 3$
Therefore $2(x-1)\times 3=-1$
$\Rightarrow 6(x-1)=-1$
$6x-6=-1$
$6x=5$
$x=\large\frac{5}{6}$
Step 3:
When $x=\large\frac{5}{6}$
$y=\big(\large\frac{5}{6}\big)$$-2\big(\large\frac{5}{6}\big)$$+7$
$y=\large\frac{25}{36}-\frac{10}{6}$$+7$
$\;\;=\large\frac{25-60+252}{36}$
$\;\;=\large\frac{217}{36}$
Hence the points are $(\large\frac{5}{6},\frac{217}{36})$
Step 4:
Equation of the tangent which is perpendicular to $5y-15x=13$ at $\big(\large\frac{5}{6},\frac{217}{36}\big)$ is
$y-\large\frac{217}{36}=\frac{-1}{3}$$(x-\large\frac{5}{6})$
On simplifying we get,
$36y-217=-12(x-\large\frac{5}{6})$
$36y-217=-12x+10$
$\Rightarrow 12x+36y-227=0$
This is the required equation of the tangent.
answered Jul 11, 2013 by sreemathi.v
 

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