QUESTION IMAGE
Question
select all the points which lie on the parabola with focus at (0, 2) and directrix at y = -2.
a. (-4,2)
b. (0,0)
c. (0,2)
d. (4,3)
e. (8,8)
Step1: Recall the definition of a parabola
A parabola is the set of all points \((x, y)\) such that the distance from the point to the focus is equal to the distance from the point to the directrix. The focus is \((0, 2)\) and the directrix is \(y=-2\). The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), and the distance from a point \((x,y)\) to the directrix \(y = k\) is \(|y - k|\). So for a point \((x,y)\) on the parabola, \(\sqrt{(x - 0)^2+(y - 2)^2}=|y+ 2|\) (since distance to directrix \(y=-2\) is \(|y-(-2)|=|y + 2|\)).
Step2: Square both sides to eliminate square root and absolute value
Squaring both sides of \(\sqrt{x^{2}+(y - 2)^{2}}=|y + 2|\), we get \(x^{2}+(y - 2)^{2}=(y + 2)^{2}\).
Step3: Expand the equations
Expand \((y - 2)^{2}=y^{2}-4y + 4\) and \((y + 2)^{2}=y^{2}+4y + 4\). Substitute into the equation:
\(x^{2}+y^{2}-4y + 4=y^{2}+4y + 4\)
Step4: Simplify the equation
Subtract \(y^{2}+ 4\) from both sides: \(x^{2}-4y=4y\). Then add \(4y\) to both sides: \(x^{2}=8y\), or \(y=\frac{1}{8}x^{2}\). Now we will check each point:
Check point A: \((-4,2)\)
Substitute \(x=-4\), \(y = 2\) into \(y=\frac{1}{8}x^{2}\). RHS: \(\frac{1}{8}\times(-4)^{2}=\frac{1}{8}\times16 = 2\). LHS: \(2\). So \(2 = 2\), the point lies on the parabola.
Check point B: \((0,0)\)
Substitute \(x = 0\), \(y=0\) into \(y=\frac{1}{8}x^{2}\). RHS: \(\frac{1}{8}\times0^{2}=0\). LHS: \(0\). So \(0=0\), the point lies on the parabola.
Check point C: \((0,2)\)
Substitute \(x = 0\), \(y = 2\) into \(y=\frac{1}{8}x^{2}\). RHS: \(\frac{1}{8}\times0^{2}=0\). LHS: \(2\). \(0
eq2\), so the point does not lie on the parabola.
Check point D: \((4,3)\)
Substitute \(x = 4\), \(y = 3\) into \(y=\frac{1}{8}x^{2}\). RHS: \(\frac{1}{8}\times4^{2}=\frac{1}{8}\times16 = 2\). LHS: \(3\). \(2
eq3\), so the point does not lie on the parabola.
Check point E: \((8,8)\)
Substitute \(x = 8\), \(y = 8\) into \(y=\frac{1}{8}x^{2}\). RHS: \(\frac{1}{8}\times8^{2}=\frac{1}{8}\times64 = 8\). LHS: \(8\). So \(8 = 8\), the point lies on the parabola.
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A. \((-4,2)\), B. \((0,0)\), E. \((8,8)\)