equation of tangent

**mastermin346** · October 11th 2010, 04:37 AM

The normal to the curve $y=x^2 - 6x + 1$ at the point $p$ is parallel to the straight line.Find the equation of the $2y = x + 3$ tangent to the curve at the point $p$

**Archie Meade** · October 11th 2010, 05:34 AM

Originally Posted by mastermin346

The normal to the curve $y=x^2 - 6x + 1$ at the point $p$ is parallel to the straight line $2y = x + 3$ .

Find the equation of the tangent to the curve at the point $p$

If this is how your post should have read, then calculating the derivative of $x^2-6x+1$

will give you the slope of the tangent.

The slope of the line $2y=x+3\Rightarrow\ y=\frac{1}{2}x+\frac{3}{2}$
is $\frac{1}{2}$

The normal and the tangent are perpendicular, so you can find what the tangent slope needs to be.

The derivative being equal to this tangent slope will obtain the x co-ordinate of $p$

The y co-ordinate is obtained by placing the x co-ordinate into f(x).

Finally, to get the equation of the tangent, use the equation of the line formula,
using the slope calculated and the co-ordinates of $p$

**mastermin346** · October 11th 2010, 06:14 AM

i get:

$y=x^2 -3x + 5$

$\frac{dy}{dx} = 2x - 6$

$2y = x + 3$

$y = \frac{1}{2}x + \frac{3}{2}$

$(m_1)(m_2) = -1$

$\frac{1}{2}(m_2) = -1$

$m_2 = -2$

$2x - 6 = -2$

$x = 2$

when $x = 2$

$y = 2^2 - 6(2) + 1$

$y = -7$

$using (2,-7)$ and $m = -2$

$y + 7 = -2 (x - 2)$

$y + 7 = -2x + 4$

is that correct sir?

**Archie Meade** · October 11th 2010, 06:22 AM

Originally Posted by mastermin346

i get:

$y=x^2 -6x + 1$

$\frac{dy}{dx} = 2x - 6$

$2y = x + 3$

$y = \frac{1}{2}x + \frac{3}{2}$

$(m_1)(m_2) = -1$

$\frac{1}{2}(m_2) = -1$

$m_2 = -2$

$2x - 6 = -2$

$x = 2$

when $x = 2$

$y = 2^2 - 6(2) + 1$

$y = -7$

$using (2,-7)$ and $m = -2$

$y + 7 = -2 (x - 2)$

$y + 7 = -2x + 4$

is that correct sir?

Yes, that's it. (small typo on your first line)

**emakarov** · October 11th 2010, 08:33 AM

Originally Posted by mastermin346

The normal to the curve $y=x^2 - 6x + 1$ at the point $p$ is parallel to the straight line.Find the equation of the $2y = x + 3$ tangent to the curve at the point $p$

Shouldn't it say the following?

The normal to the curve $y=x^2 - 6x + 1$ at the point $p$ is parallel to the straight line $2y = x + 3$ . Find the equation of the tangent to the curve at the point $p$ .

Did you copy-paste from a two-column PDF file?

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