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Let $y^2=\dfrac{x^2(a-x)}{(a+x)}$ be given.

I am not quite sure how the right spheroid works in general. I have calculated the point on the curve by plugging the random value of $x = 1$ and got $y$ approximately $0.64565$, hence the title.

I am not quite sure how to calculate the tangent and normal line of the right spheroid, and what is the actual formula for the right spheroid in general?

Thanks in advance!

TShiong
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Crouch Potato
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  • A spheroid is a type of surface, not a specific surface. You need the defining characteristics of the specific spheroid you are speaking of. Check here for a definition. – John Wayland Bales May 21 '23 at 15:59
  • This is what I have as a problem I am working on. Determine the coordinates of some point on the right spheroid if the parameter a=2.5. For that point, determine the tangent and normal line at that point. – Crouch Potato May 21 '23 at 16:08
  • Are you working in 2D or 3D? And please do define what you mean by spheroid? And what has $a$ to do about it all? – MasB May 21 '23 at 16:11
  • By right spheroid I mean this curve : y^2=(x^2 (a-x))/(a+x), if I understood the concept correctly from this site : https://mathshistory.st-andrews.ac.uk/Curves/Right/curvesapplet/ – Crouch Potato May 21 '23 at 16:14
  • You must find the implicit derivative of $y$ with respect to $x$ using the equation of the curve and substitute the values of $a,x,y$ at the point of tangency to find the slope of the tangent line. Then you have the coordinates of a point on the tangent line and its slope, which is all you need to find the equation. The slope of the normal line is the negative reciprocal of the slope of the tangent line, so you can also find the equation of the normal line. – John Wayland Bales May 21 '23 at 18:30
  • Please read and copy carefully. The article you refer to in mathshistory is about a strophoid, not a spheroid. – MasB May 21 '23 at 21:01

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I have decided to make my comment an answer.

The operations involved should be in your skill set.

You must find the implicit derivative of with respect to $x$ using the equation of the curve and substitute the values of $a,x,y$ at the point of tangency to find the slope of the tangent line. Then you have the coordinates of a point on the tangent line and its slope, which is all you need to find the equation. The slope of the normal line is the negative reciprocal of the slope of the tangent line, so you can also find the equation of the normal line.

Note that there are two points on the curve with $x$-coordinate $1$.

Image of curve with tangent & normal lines

John Wayland Bales
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