Find the equation of tangent and normal to the curve

**zorro** · December 28th 2009, 09:58 PM

Question : Find the eq of tangent and normal to the curve
$y(x-2)(x-3)-x+7 = 0$
at the point where the curve meets the x-axis
-----------------------------------------------------------------
Solution:

$y = \frac{x-7}{x^2 - 5x +6 }$ .............Is that correct?

$\therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}$ ...................Is that correct?????

**mr fantastic** · December 28th 2009, 10:38 PM

Originally Posted by zorro

Question : Find the eq of tangent and normal to the curve
$y(x-2)(x-3)-x+7 = 0$
at the point where the curve meets the x-axis
-----------------------------------------------------------------
Solution:

$y = \frac{x-7}{x^2 - 5x +6 }$ .............Is that correct?

$\therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}$ ...................Is that correct?????

differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.

**zorro** · December 29th 2009, 02:42 PM

Originally Posted by mr fantastic

differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.

$\frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}$
what should i do??????
Now eg of tangent is at $(x_0,y_0)$ is given by

$(y-y_0) = \frac{dy}{dx} (x-x_0)$

but i dont know $(x_0,y_0)$ .................what should i do??????

**mr fantastic** · December 29th 2009, 04:02 PM

Originally Posted by zorro

$\frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}$
what should i do??????
Now eg of tangent is at $(x_0,y_0)$ is given by

$(y-y_0) = \frac{dy}{dx} (x-x_0)$

but i dont know $(x_0,y_0)$ .................what should i do??????

When the curve meets the x-axis, y = 0. Therefore $y_0 = 0$ . To get $x_0$ , solve $-x + 7 = 0$ (I hope you see why).

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