Find the focus and directrix of the parabola y = .5(x + 2)2 – 3

Mathematics

1 answer:

Ray Of Light [21]2 years ago

8 0

Answer:

<h3>comparing equation with standard equation x </h3><h3>2</h3><h3> =4ay</h3><h3>a=2/5</h3><h3>co-ordinates of focus =(o,a) i.e. (0, 2/5)</h3><h3>equation of directrix=y=-a i.e. y=-(2/5)</h3><h3>length of latus rectum= 4a i.e. 8/5</h3><h3>co-ordinates of latus rectum=(-2a,a) and (2a,a) i.e. (-4/5,2/5) and (4/5,2/5)..</h3>

Step-by-step explanation:

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Gekata [30.6K]

Answer:

Converting the equation $x^2-20x+13=0$ into completing the square method we get: $\mathbf{(x-10)^2=87}$

Step-by-step explanation:

we are given quadratic equation: $x^2-20x+13=0$

And we need to convert it into completing the square method.

Completing the square method is of form: $a^2-2ab+b^2=(a-b)^2$

Looking at the given equation $x^2-20x+13=0$

We have a = x

then we have middle term 20x that can be written in form of 2ab So, we have a=x and b=? Multiplying 10 with 2 we get 20 so, we can say that b = 20

So, 20x in form of 2ab can be written as: 2(x)(10)

So, we need to add and subtract (10)^2 on both sides

$x^2-20x+13=0\\x^2-2(x)(10)+(10)^2-(10)^2+13=0\\(x^2-2(x)(10)+(10)^2) \:can\: be\: written\: as\: (x-10)^2 \\(x-10)^2-100+13=0\\(x-10)^2-87=0\\(x-10)^2=87$