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find an equation for the parabola with focus $(0,-4)$ and directrix $y …

Question

find an equation for the parabola with focus $(0,-4)$ and directrix $y = 4$.

Explanation:

Step1: Recall the definition of parabola

A parabola is the set of points \((x,y)\) such that the distance from \((x,y)\) to the focus \((0,-4)\) is equal to the distance from \((x,y)\) to the directrix \(y = 4\).
The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
The distance from \((x,y)\) to \((0,-4)\) is \(\sqrt{(x - 0)^2+(y+4)^2}\), and the distance from \((x,y)\) to the line \(y = 4\) is \(|y - 4|\).
So, \(\sqrt{x^{2}+(y + 4)^{2}}=|y - 4|\).

Step2: Square both sides

Squaring both sides of the equation \(\sqrt{x^{2}+(y + 4)^{2}}=|y - 4|\), we get \(x^{2}+(y + 4)^{2}=(y - 4)^{2}\).
Expand the squares: \(x^{2}+y^{2}+8y + 16=y^{2}-8y + 16\).

Step3: Simplify the equation

Subtract \(y^{2}+16\) from both sides: \(x^{2}+y^{2}+8y + 16-(y^{2}+16)=y^{2}-8y + 16-(y^{2}+16)\).
This simplifies to \(x^{2}+8y=-8y\).
Then, \(x^{2}=-16y\).

Answer:

\(x^{2}=-16y\)