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It is generally told to us students to mug up this method for find the point of intersection of the pair of straight lines given by: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ Suppose $\Phi = ax^2 + 2hxy + by^2 + 2gx + 2fy + c$, then by partially differentiating $\Phi$ with respect to $x$ and again with respect to $y$ gives the two equations: $$ \dfrac{\partial\Phi}{\partial x} = 2(ax + hy + g) \\[2ex] \dfrac{\partial\Phi}{\partial y} = 2(hx + by + f)$$ Setting them both equal to zero and solving them gives the point of intersection of the two lines represented by the equation $\Phi = 0$. So why does this happen? Is there an explanation for it? Or atleast an intuitive explanation? (Please include the intuitive explanation also! It would help heaps.) I do not know a lot of geometry, this is the first time I am seeing this equation or the chapter, so if possible, please give the complete explanation.

That's a lot of asking from my side, so thank you very much if you decide to answer!

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  • $\begingroup$ The straigt lines have no geometrical interpretation. You just calculate the partial derivatives and set them equal to zero. In case of x you look for a point where slope in x-direction is equal to zero. Similar for the partial derivative w.r.t y. Thus in both directions you have a slope of zero. And then you have three possible cases: relative maximum, relative minimum or saddle point. Make a sketch. $\endgroup$ – callculus Jul 27 '16 at 6:45
  • $\begingroup$ @callculus I don't know what a saddle point is... that is what i meant when i said i don't have a lot of knowledge on this! I am studying in class 12 only right now! $\endgroup$ – FreezingFire Jul 28 '16 at 4:42
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Here's a graphical, hand-wavy, interpretation:

$\Phi(x,y)=0$ is the implicit function of a certain graph in the $(x,y)$ plane. We know that: $$\frac{dy}{dx}=-\frac{\dfrac{\partial \Phi}{\partial x}}{\dfrac{\partial \Phi}{\partial y}}$$

If $\dfrac{\partial \Phi}{\partial x}=0$ and $\dfrac{\partial \Phi}{\partial y}=0$ then $dy/dx$ is indeterminate, and the only place that's true for two lines is their intersection.

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  • $\begingroup$ That is a nice interpretation! I will accept this if i get no better answer...Thank you for your answer! $\endgroup$ – FreezingFire Jul 28 '16 at 4:43
  • $\begingroup$ I don't think that's a correct interpretation, first phi is identically zero, so there is no meaning to taking it's derivative, moreover x and y aren't independent of each other. Next 0/0 isn't the only indeterminate form available, so why choose that. Also how can you claim that an indeterminate slope results in a crossover of the lines. In most cases(say a lemniscate curve) forcing a derivative on the intersection point, gives us both the slopes of each individual smooth curve. $\endgroup$ – HyperBean Feb 11 '19 at 6:17
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The function $\Phi$ is called a quadric and describes two-dimensional analogues of conic sections (ellipse, hyperbola, &c). Setting $\Phi$ equal to some constant value is equivalent to taking a horizontal slice through the surface. The resulting curve is called a “level curve” of the surface. It turns out that the only quadric surface that can have a pair of intersecting lines as a level curve is what’s called a hyperbolic paraboloid, which looks kind of like a Pringles potato chip or a saddle. You can see an example of it and other quadrics in this Wikipedia article. You get this surface when the coefficients of $\Phi$ satisfy certain relationships that aren’t important here.

If you take horizontal slices through this surface, you’ll get hyperbolas. If the slice goes through the surface’s saddle point—the spot where it flattens out—this hyperbola is degenerate: it’s a pair of straight lines that intersect at the saddle point. In analytical terms, a nicely-behaved surface “flattening out” at a point means that all partial derivatives of the function that defines the surface are zero there—it’s locally flat in every direction. This condition can be packaged up into a single equation by using the function’s gradient: $\nabla\Phi=0$.

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At the pictures below you can see a (local) maximum, a (local) minimum and a saddle point. The generic term for these three points is stationary point. They all have in common that the slope in $x$ direction and the slope in $y$ direction are equal to zero.

In mathematical terms they are the partial derivative w.r.t $x$ and $y$ respectively which are set equal to zero. Let the function with two independent variables be $z(x,y)$, then we have

$\frac{\partial z}{\partial x} =0$

$\frac{\partial z}{\partial y} =0$

These two equations are the conditions for a stationary point.

The special about the saddle point is that it has a local maximum in x direction and a local minimum in y direction et vice versa.

enter image description here

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$$ L=q.r=ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\\ L_x=rq_x+qr_x=ax+hy+g\\ L_y=rq_y+qr_y=hx+by+f\\ $$ When $q=0,r=0\implies L_x=0,L_y=0\implies$ point of intersection of $q=0,r=0$ is the same as the point of intersection of $L_x=0,L_y=0$

$q=0,r=0$ whenever $L_x=0,L_y=0$

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