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Find the coordinates of the point where the tangent to the curve $y=x^2+1$ at the point $(2,5)$ meets the normal to the same curve at the point $(1,2).$

I tried to form 2 equations for each set of coordinates given, then solve then simultaneously to get the 'y and x' coordinates needed, however I didn't seem to get the answer right.

amWhy
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user108815
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2 Answers2

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  • First, you need to find $f'(x)= 2x$.
  • Then to evaluate the slope of the tangent line at the point $(2, 5)$, $m_1 = f'(2) = 4$. With slope $m_1=4$, and the point $(x_0, y_0) = (2, 5)$, use the slope-point form of an equation to obtain the equation of that tangent line.

$(y - y_0) = m_1(x - x_0)\tag{Point-Slope Equation Of Line}$

  • To find the slope of the normal line at the point $(1, 2)$, first evaluate $f'(1) = m_2= 2$, the slope of the tangent line at that point. Then the slope of the normal line to the curve at that same point is $m_\perp = -\dfrac 1{m_2} = -\dfrac 12$. Using the slope of the normal line $m_\perp = -\dfrac 12$ and the point $(x_0, y_0) = (1, 2)$, use the slope point form of an equation to obtain the equation of the line normal to the curve at $(1, 2)$:
    $$(y - y_0) = m_\perp(x - x_0)$$

  • Finally, find where the tangent line and the normal line intersect by setting the equations equal to one another. Solve for $x$, then find the corresponding $y$ value using the equation of either line.

amWhy
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$y'(x) = 2x$

$y'(2) = 4$, so $4$ is the slope of the tangent at $(2,5)$.

You have a point and a slope, so you should be able to find the equation of a line.

Then $y'(1) =2$, so we invert and negate this to get the normal slope ($=-1/2$).

Point + slope --> equation.

Etc.

The Chaz 2.0
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