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Area between Two Curves

The slope of ...

Question

The slope of normal to curve $y = f (x)$ , where $x = t^{2} - 5 t + 4$ and $y = t^{2} - 7 t + 10$ at $(- 2, - 2),$ is

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\frac{1}{2}

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Solution

The correct option is C $1$
Given curve : $x = t^{2} - 5 t + 4$
On differentiating both sides w.r.t. $t$
$\frac{d x}{d t} = 2 t - 5$
and $y = t^{2} - 7 t + 10$
On differentiating on both sides w.r.t. $t$
$\frac{d y}{d t} = 2 t - 7$
For $x = - 2$
$t^{2} - 5 t + 4 = - 2$
$\Rightarrow t^{2} - 5 t + 6 = 0$
$\Rightarrow t = 2, 3$
and For $y = - 2$
$\Rightarrow t^{2} - 7 t + 12 = 0$
$\Rightarrow t = 4, 3$
So, $t = 3$ at point $(- 2, - 2)$
Now,
$\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x} = \frac{2 t - 7}{2 t - 5}$
Slope of tangent at $t = 3$
$\frac{d y}{d x} {∣ ∣ ∣}_{(t = 3)} = - 1$
$∴$ slope of normal $= 1$

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