What are the normals of a parabola passing through a given point?

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Find all the normals of function $$y = 2x^2 + 4x + \tfrac{7}{4}$$ which goes through the point $$\left( 3, \tfrac{15}{2} \right)$$.
 
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Why? If you are posting this because you want help with it then you should show us what you do understand about it yourself so that we will know what kinds of hints and help you need.

Do you understand what a "normal" to a graph is? Do you understand that a normal to a graph, at a point, is perpendicular to the tangent to that graph at that point? Can you find the tangent to $y= 2x^2+ 4x+ \frac{7}{4}$.

Notice that $2(3)^2+ 4(3)+ \frac{7}{2}= 18+ 12+ \frac{7}{4}= 30+ \frac{7}{4}= \frac{127}{4}$ not $\frac{15}{2}$ so the given point is not on the curve. You will need to find the tangent line at some point $\left(a, 2a^2+ 4a+ \frac{7}{2}\right)$ then find a so that tangent line passes through (3, 15/4). Once you have found the point and the equation of the tangent, it should be easy to find the normal line.
 
HallsofIvy said:
Why? If you are posting this because you want help with it then you should show us what you do understand about it yourself so that we will know what kinds of hints and help you need.

Do you understand what a "normal" to a graph is? Do you understand that a normal to a graph, at a point, is perpendicular to the tangent to that graph at that point? Can you find the tangent to $y= 2x^2+ 4x+ \frac{7}{4}$.

Notice that $2(3)^2+ 4(3)+ \frac{7}{2}= 18+ 12+ \frac{7}{4}= 30+ \frac{7}{4}= \frac{127}{4}$ not $\frac{15}{2}$ so the given point is not on the curve. You will need to find the tangent line at some point $\left(a, 2a^2+ 4a+ \frac{7}{2}\right)$ then find a so that tangent line passes through (3, 15/4). Once you have found the point and the equation of the tangent, it should be easy to find the normal line.

This thread was posted in our "Challenge Questions and Puzzles" forum, so that means the OP has the solution and finds the problem interesting and so wishes to post the problem as a challenge to the community. (Yes)

By the way, I edited your post to hide anything that might give anything away for those who don't wish to see any hints before attempting to solve it themselves.