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I've been asked to prove that the normal to the curve $y=2x^2 - 3x^{-1/2}$ at the point $(1,-1)$ passes through the point $(12,3)$.

$\frac{dy}{dx} = 4x + \frac{3}{2}x^{-3/2}$

Hence, at the point $(1,-1)$, the gradient of the tangent $=\frac{11}{2}$

Therefore, the gradient of the normal at that point$=-\frac{2}{11}$

So, the equation for the normal is $$y+1=-\frac{2}{11}(x-1)$$

$$11y=-2x-9$$

However, when I substitute the value $12$ in for $x$ I get $y=-3$, rather than $3$, so that equation can't be right.

What have I done wrong?

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    $\begingroup$ Looking at that graph using a graphing calculator leads me to believe that you are correct and the question is wrong. $\endgroup$
    – imulsion
    May 13, 2015 at 20:59
  • $\begingroup$ I am lead to that conclusion @SufyanNaeem, although I can't see how it is wrong $\endgroup$
    – Charon
    May 13, 2015 at 21:02
  • $\begingroup$ Thanks @imulsion - I thought that might be the case. Certainly, I can't find a flaw in my logic. $\endgroup$
    – Charon
    May 13, 2015 at 21:02
  • $\begingroup$ @SufyanNaeem - I take it that you mean therefore that the initial quadratic is wrong? $\endgroup$
    – Charon
    May 13, 2015 at 21:06

1 Answer 1

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You are correct, the question is wrong (or printed incorrectly). Look at this graph:

enter image description here

The red line is the graph of the equation in the question, the green line is the graph of the equation of the normal. It is clear that your normal line is correct for this graph.

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    $\begingroup$ Red line is not "an equation" nor green one is. They are expressing their respective equations. Be careful with terminologies. $\endgroup$ May 13, 2015 at 21:06
  • $\begingroup$ @SufyanNaeem If you want to be pedantic, I'll change my wording $\endgroup$
    – imulsion
    May 13, 2015 at 21:07
  • $\begingroup$ This is not a matter of being pedantic or not but it is the matter of right or wrong. You should be formal while writing posts in Math.SE. :) $\endgroup$ May 13, 2015 at 21:09

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