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Prove that the equation y^(2)+2ax+2by+c=...

Prove that the equation `y^(2)+2ax+2by+c=0` represents a parabola whose axis is parallel to the axis of x. Find its vertex.

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To prove that the equation \( y^2 + 2ax + 2by + c = 0 \) represents a parabola whose axis is parallel to the x-axis and to find its vertex, we will follow these steps: ### Step 1: Identify the general form of a conic section The general form of a conic section is given by: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \] For our equation \( y^2 + 2ax + 2by + c = 0 \), we can identify the coefficients: - \( A = 0 \) - \( B = 0 \) - \( C = 1 \) - \( D = 2a \) - \( E = 2b \) - \( F = c \) ### Step 2: Calculate the discriminant The discriminant \( \Delta \) for conic sections is given by: \[ \Delta = B^2 - 4AC \] Substituting the values we identified: \[ \Delta = 0^2 - 4(0)(1) = 0 \] Since \( \Delta = 0 \), this indicates that the equation represents a parabola. ### Step 3: Check the condition for a parabola For the equation to represent a parabola, we also check the condition: \[ h^2 - ab = 0 \] Where \( h = \frac{D}{2} \) and \( a = A \), \( b = C \). Here: - \( h = \frac{2a}{2} = a \) - \( a = 0 \) - \( b = 1 \) Calculating \( h^2 - ab \): \[ h^2 - ab = a^2 - (0)(1) = a^2 \] Since \( a^2 = 0 \), this condition is satisfied. ### Step 4: Rearranging the equation Now, we rearrange the original equation to find the vertex: \[ y^2 + 2by = -2ax - c \] Completing the square for the \( y \) terms: \[ y^2 + 2by + b^2 = -2ax - c + b^2 \] This can be rewritten as: \[ (y + b)^2 = -2ax - c + b^2 \] Thus, we have: \[ (y + b)^2 = -2a\left(x + \frac{c - b^2}{2a}\right) \] ### Step 5: Identify the vertex From the equation \( (y + b)^2 = -2a\left(x + \frac{c - b^2}{2a}\right) \), we can identify the vertex of the parabola: - The vertex \( (h, k) \) is given by: \[ \left(-\frac{c - b^2}{2a}, -b\right) \] ### Conclusion Thus, we have proved that the equation \( y^2 + 2ax + 2by + c = 0 \) represents a parabola whose axis is parallel to the x-axis, and the vertex is: \[ \left(-\frac{c - b^2}{2a}, -b\right) \]
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ICSE-PARABOLA-EXERCISE 23
  1. The focus at (-2,-1) and the latus rectum joins the points (-2,2) and ...

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  2. Find the equation of a parabola whose vertex at (-2,3) and the focus a...

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  3. Find the equation of parabola if it's vertex is at (0,0) and the focu...

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  4. Find the equation of the parabola whose vertex is at (0,0) and the foc...

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  5. The axis parallel to the x-axis, and the parabola passes through (3,3)...

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  6. The axis parallel to the x-axis, and the parabola passes through the p...

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  7. The parabola y^2=4px passes thrugh the point (3,-2). Obtain the length...

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  8. Prove that the equation y^(2)+2ax+2by+c=0 represents a parabola whose ...

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  9. Of the parabola, 4(y-1)^(2)= -7(x-3) find The length of the latus re...

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  10. Of the parabola, 4(y-1)^(2)= -7(x-3) find The coordinates of the foc...

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  11. Find the vertex, focus, and directrix of the following parabolas: y^...

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  12. Find the vertex, focus, and directrix of the following parabolas: x^...

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  13. Find the vertex, focus and directix of the parabola (x-h)^(2)+4a(y-k)=...

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  14. Find the equatin to the parabola whose axis is parallel to the y-xis a...

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  15. Find the coordinates of the point on the parabola y^(2)=8x whose focal...

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  16. If the ordinate of a point on the parabola y^(2)=4ax is twice the latu...

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  17. Find the equation of the parabola whose focus is at the origin, and wh...

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  18. The directrix of a conic section is the straight line 3x-4y+5-0 and th...

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  19. Find the equation to the parabola whose focus is (-2,1) and directrix ...

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  20. The length of the latus rectum of the parabola whose focus is (3,3) an...

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