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 (|x+2y|)/(√(1²+2²)). Squaring both sides, we get (x-3)² + (y-6)² = ((x+2y)²)/(5).

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 (|x+2y|)/(√(1²+2²)). Squaring both sides, we get (x-3)² + (y-6)² = ((x+2y)²)/(5).

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.

By comparing the derived equation 4x² + y² - 4xy - 30x - 60y + 225 = 0 with the given general form α x² + β y² - xy - 30x - 60y + 225 = 0, we can identify the coefficients: α = 4, β = 1, and = 4. Finally, we calculate their sum: 4 + 1 + 4 = 9.

By comparing the derived equation 4x² + y² - 4xy - 30x - 60y + 225 = 0 with the given general form α x² + β y² - xy - 30x - 60y + 225 = 0, we can identify the coefficients: α = 4, β = 1, and = 4. Finally, we calculate their sum: 4 + 1 + 4 = 9.

If the equation of the parabola with vertex $\mathbf{V}\left(\frac{3}{2}, 3\right)$ and the directrix $x + 2y = 0$ is $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$ , then $\alpha + \beta + \gamma$ is equal to:

Answer: 9

Step-by-step solution

Step 1: Find the focus of the parabola

The axis of the parabola passes through the vertex $V\left(\frac{3}{2}, 3\right)$ and is perpendicular to the directrix $x + 2y = 0$ . The slope of the directrix is $m_d = -\frac{1}{2}$ , so the slope of the axis is $m_a = 2$ . The equation of the axis is $y - 3 = 2\left(x - \frac{3}{2}\right)$ , 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 $x+2(2x)=0 \implies 5x=0 \implies x=0$ . So $y=0$ . Thus, $K=(0{,}0)$ . Since the vertex $V$ is the midpoint of the focus $F(x_1, y_1)$ and the point $K(0{,}0)$ on the directrix, we have $\frac{x_1+0}{2} = \frac{3}{2}$ and $\frac{y_1+0}{2} = 3$ . This gives $x_1=3$ and $y_1=6$ . 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 $\sqrt{(x-3)^2 + (y-6)^2}$ . The perpendicular distance from $P(x,y)$ to the directrix $x+2y=0$ is $\frac{|x+2y|}{\sqrt{1^2+2^2}}$ . Squaring both sides, we get $(x-3)^2 + (y-6)^2 = \frac{(x+2y)^2}{5}$ .

Step 3: Expand and simplify the equation

Expand the squared terms on both sides of the equation. On the left side, we have $x^2 - 6x + 9 + y^2 - 12y + 36$ . On the right side, $(x+2y)^2$ expands to $x^2 + 4xy + 4y^2$ . Multiply the left side by 5.

Step 4: Rearrange into the general form

Distribute the 5 on the left side: $5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2$ . Combine like terms and move all terms to one side to match the given general form $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$ . This gives $4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$ .

Step 5: Identify coefficients and calculate sum

By comparing the derived equation $4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$ with the given general form $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$ , we can identify the coefficients: $\alpha = 4$ , $\beta = 1$ , and $\gamma = 4$ . Finally, we calculate their sum: $4 + 1 + 4 = 9$ .

More from this chapter

All Conic Sections solutions →

Solve your own maths question free →