If the equation of the parabola with vertex V(23,3) and the directrix x+2y=0 is αx2+βy2−γxy−30x−60y+225=0, then α+β+γ is equal to:
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Step-by-Step Solution
Step 1: Find the focus of the parabola
The axis of the parabola passes through the vertex V(23,3) and is perpendicular to the directrix x+2y=0. The slope of the directrix is md=−21, so the slope of the axis is ma=2. The equation of the axis is y−3=2(x−23), which simplifies to y=2x. The intersection of the axis and the directrix is point K. Substituting y=2x into x+2y=0, we get x+2(2x)=0⟹5x=0⟹x=0. So y=0. Thus, K=(0,0). Since the vertex V is the midpoint of the focus F(x1,y1) and the point K(0,0) on the directrix, we have 2x1+0=23 and 2y1+0=3. This gives x1=3 and y1=6. So the focus is F(3,6).
Step 2: Derive the equation of the parabola
For any point P(x,y) on the parabola, its distance from the focus F(3,6) is equal to its perpendicular distance from the directrix x+2y=0. The distance from P(x,y) to F(3,6) is (x−3)2+(y−6)2. The perpendicular distance from P(x,y) to the directrix x+2y=0 is 12+22∣x+2y∣. Squaring both sides, we get (x−3)2+(y−6)2=5(x+2y)2.
Step 3: Expand and simplify the equation
Expand the squared terms on both sides of the equation. On the left side, we have x2−6x+9+y2−12y+36. On the right side, (x+2y)2 expands to x2+4xy+4y2. Multiply the left side by 5.
Step 4: Rearrange into the general form
Distribute the 5 on the left side: 5x2−30x+45+5y2−60y+180=x2+4xy+4y2. Combine like terms and move all terms to one side to match the given general form αx2+βy2−γxy−30x−60y+225=0. This gives 4x2+y2−4xy−30x−60y+225=0.
Step 5: Identify coefficients and calculate sum
By comparing the derived equation 4x2+y2−4xy−30x−60y+225=0 with the given general form αx2+βy2−γxy−30x−60y+225=0, we can identify the coefficients: α=4, β=1, and γ=4. Finally, we calculate their sum: 4+1+4=9.