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Find $\frac{dy}{dx}$ of $x^2 +y^2 +xy=27$ and use it to find the equations of the tangent and normal lines to the curve $(3\sqrt{3}, 0)$.

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3 Answers 3

To find $\dfrac{dy}{dx}$:

$$d(x^2 + y^2 + xy) = 2x dx + 2y dy + xdy + ydx = d(27) = 0\tag{1}$$ $$(x + 2y)\,dy = -(2x + y)\,dx$$

Then we can express $$ \frac{dy}{dx} = -\frac{2x + y}{x+2y}\tag{2}$$

Now, using your given point $(3\sqrt 3, 0)$, evaluate $\large\frac{dy}{dx}$ at the given $x$ and $y$ coordinate in order to:

  • determine the slope of the line tangent to the ellipse at that point, and
  • then from that slope, determine the slope of the line normal to the ellipse$^+$ at that point.
  • Then, use the point-slope form of the equation of a line$^{++}$ to determine the equations of each of the line tangent to the ellipse at $(3\sqrt 3,0)$, and the line normal to the ellipse at that point.

$^+$Recall: if slope of the tangent line is $\;\large\frac ab\;$, then the slope of the line normal at that point is $\;\large-\frac ba$.
$^{++}$ Recall: the point-slope of the equation of a line, given point $\,(x_0, y_0)\,$ and slope $\,m,\,$ is given by $\;(y - y_0) = m(x - x_0)$

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Nice answer. I always set the relation as $F(x,y)=0$, then use the following $$y'=\frac{-F_x}{F_y}$$ instead. +1 for tomorrow. –  Babak S. Feb 3 '13 at 20:05
    
user59633: if you're still struggling, let me know if you have any questions I can address. –  amWhy Feb 3 '13 at 22:49

$d(x^2 + y^2 + xy) = 2x dx + 2y dy + xdy + ydx = d(27) = 0$.

Then $$ \frac{dy}{dx} = -\frac{2x + y}{x+2y}$$

Then with this, you can find the slope of the tangent line at $(3\sqrt 3,0)$, then find the normal.

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I am confused as to exactly how to find the tangent and normal lines? Could you provide a rough idea about how to do so? –  user59633 Feb 3 '13 at 19:35

There's a much simpler way to find the tangent line at a point on an ellipse (or any conic section). If the conic has equation $ax^2 + 2bxy + cy^2 + 2dx + 2ey + f =0$, then the tangent at the point $(x_0, y_0)$ is $ax_0x + b(x_0y + xy_0) + cy_0y + d(x+x_0) + e(y+y_0) + f = 0$.

Informally, you "half-way substitute" $(x_0, y_0)$ in place of $(x, y)$.

Generally, the line you get this way is the "polar line" through the point $(x_0, y_0)$. But, if the point lies on the conic, then the polar line is a tangent.

Once you have the tangent line, getting the normal line is easy, of course.

No need for any differentiation.

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