Solve Tangent Equation: Find A, B, & C for Ellipse

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To find constants A, B, and C for the ellipse equation 4x^2 + y^2 + Ax + By + C = 0, it must pass through the origin, indicating that C = 0. The ellipse is also tangent to the x-axis at the origin, which means the gradient at that point must have no x-component, allowing for the determination of A. By substituting the point P(-1, 2) into the equation, the value of B can be calculated. The discussion emphasizes using the gradient to ensure the ellipse's properties are satisfied. Ultimately, the values of A, B, and C can be derived from these conditions.
kidia
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Hi any idea on this please.

Find the values of constant A,B and C if the ellipse 4x^2 +y^2+Ax+By+C=0 is to be tangent to the x-axis at the origin and to pass through the point P(-1,2)
 
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what i would do first is to put the equation into a form more readable as far as ellipces aree concerned
 
You can also do it straight ahaid:

(1) The ellipse must pass through (0,0). What does this tell you about C?

(2) The ellipse must be tangent to the x-axis too. So the normal to the ellipse can not have an x-component. Recall that the gradient of a function is perpendicular to it's constant value curves. So the normal to the ellipse at the origin is given by the gradient of 4x^2 +y^2+A x+B y+C at (0,0). This gives us A.

(3) Just plug in the point P to find the remaining unknown B.
 
Dini's theorem: g(x_0, y_0)=0 and \nabla g|_{(x_0, y_0)} \neq \vec 0, then

\exists h(x) : g(x, h(x)) = 0 and

h'(x)=\frac {g_x} {g_y}
 
I didn`t got u on getting constant Value C Timbuqtu
 
kidia said:
I didn`t got u on getting constant Value C Timbuqtu
Fill in the point (0,0) and you'll find immediately that c = 0.
If not, the ellipse wouldn't go through the origin.
 
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