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tia_tia [17]
3 years ago
14

Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1

Mathematics
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

Step-by-step explanation:

eq. of directrix is y-1=0

let (x,y) be any point the parabola.

then\sqrt{ (x-6)^2+(y-2)^2}=\frac{y-1}{(-1)*2}

squaring

x²-12x+36+y²-4y+4=y²-2y+1

x²-12 x+40-1=4y-2y

2y=x²-12x+39=x²-12x+36+3

(x-6)²=2y-3

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Answer:

4

Step-by-step explanation:

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Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

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x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

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3 years ago
A rectangle's length is seven times its width. The perimeter of the rectangle is one hundred sixty feet.
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Answer:

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4 years ago
#20 four times the sum of number. and 15 is a least20. Let X represent the number. :find ball possible values for X
gayaneshka [121]
For this case what we have to take into account is the following variable:
 x = represent the unknown number
 We now write the following inequality:
 "four times the sum of number and 15 is at least 20"
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 We clear the value of x:
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 x> = - 10
 The solution set is:
 [-10, inf)
 Answer:
 
all possible values for X are:
 
[-10, inf)
6 0
3 years ago
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