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Question: How do you find the vertex, directrix and focus of \[y + 12x - 2{x^2} = 16\]?...

How do you find the vertex, directrix and focus of y+12x2x2=16y + 12x - 2{x^2} = 16?

Explanation

Solution

The equation is of the form ax2+bx+ca{x^2} + bx + cin which the vertex is the point h, k which we can find out solving this quadratic equation and hence by applying the formula we can find the vertex, directrix and focus of the equation by their general formulas obtained and the vertex of the parabola is at equal distance between focus and the directrix.

Complete step by step solution:
y+12x2x2=16y + 12x - 2{x^2} = 16
Let us rewrite the given equation in the form of ax2+bx+ca{x^2} + bx + c
y=2x212x+16y = 2{x^2} - 12x + 16 ……………………… 1
The obtained equation can be written as
2(x26x)+162\left( {{x^2} - 6x} \right) + 16
2(x26x+9)18+162\left( {{x^2} - 6x + 9} \right) - 18 + 16
2(x3)222{\left( {x - 3} \right)^2} - 2 ……………………. 2
Comparing with vertex form of equation
y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k
(h, k) is the vertex,
Hence from the equation 2, comparing with the vertex form we get
h = 3 and k = -2.
Therefore, the vertex is at (h, k) or we can say the points (3, -2) and a = 2.
Implies that vertex is at mid-point between directrix and focus and the parabola opens upward as a>0, hence directrix is below vertex.
The distance of directrix from the vertex is
d=14ad = \dfrac{1}{{4|a|}}
d=142d = \dfrac{1}{{4 \cdot 2}}
Hence, the distance of directrix is
d=18d = \dfrac{1}{8}
Therefore, the directrix is
y=(kd)y = \left( {k - d} \right)
Substituting the values of k and d we get
y=(218)y = \left( { - 2 - \dfrac{1}{8}} \right)
y=178y = - \dfrac{{17}}{8}
And the focus is at 3 i.e.,
(2+18)\left( { - 2 + \dfrac{1}{8}} \right)
We get focus at
(3,158)\left( {3, - \dfrac{{15}}{8}} \right)
Therefore,
vertex is at (3,2)\left( {3, - 2} \right),
directrix is y=178y = - \dfrac{{17}}{8} and
the focus is at (3,158)\left( {3, - \dfrac{{15}}{8}} \right)

Formula used:
y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k
(h, k) is the vertex.
d=14ad = \dfrac{1}{{4|a|}}
d is distance of directrix.
y=(kd)y = \left( {k - d} \right)

Note:
Parabolas are commonly known as the graphs of quadratic functions. The directrix is perpendicular to the axis of symmetry of a parabola and does not touch the parabola. If the axis of symmetry of a parabola is vertical, the directrix is a horizontal line. They can also be viewed as the set of all points whose distance from a certain point (the focus) is equal to their distance from a certain line (the directrix). If we consider only parabolas that open upwards or downwards, then the directrix is a horizontal line.