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Math challenge II-A geometry chapter 3

 
 
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Math challenge II-A geometry chapter 3
by Zeyin Wu - Friday, 15 October 2021, 11:24 PM
 

Another mistake in problem 3.30:

The sample solution says sqrt[1+(k+1/2)^2]=1/2-k

Why is the RHS 1/2-k? 2-(5/2-k)=k-1/2, where 2 is the y-value of point (0,2) and 5/2-k is the directrix. So the RHS should be k-1/2.

 
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Re: Math challenge II-A geometry chapter 3
by John Lensmire - Monday, 18 October 2021, 10:34 AM
 

Good question! The algebra is a little weird here because the parabola opens downwards. Hence, the focus (which is always "inside" the parabola) is actually below the vertex and the directrix is above the vertex so the $k$ value is negative.

Therefore, the distance from $(0,2)$ to the directrix is $(5/2-k) - 2 = 1/2 - k$.

Note: Sometimes it's just easier to think of distances in absolute values, so the distance is $| 1/2 - k | = | k - 1/2| $. For this problem, note that the next step will be to square both sides of the equation, so in fact you'll get the correct answer whether you use $1/2 - k$ or $k - 1/2$.

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回复: Re: Math challenge II-A geometry chapter 3
by Zeyin Wu - Monday, 18 October 2021, 4:12 PM
 

I see. Thanks a lot!