Find the equation of parabola whose focus at (-3,0)the directrix X+5= 0 ?

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Jim G. Share
Aug 21, 2017

Answer:

#y^2-4x-16=0#

Explanation:

#"from any point "(x,y)" on the parabola"#

#"the distance to the focus and the directrix from this point"#
#"are equal"#

#color(blue)"using the distance formula then"#

#sqrt((x+3)^2+(y-0)^2)=|x+5|#

#color(blue)"squaring both sides"#

#(x+3)^2+y^2=(x+5)^2#

#rArrx^2+6x+9+y^2=x^2+10x+25#

#rArry^2-4x-16=0" is the equation"#

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Write your answer here...
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Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

8
Aug 21, 2017

Answer:

#y^2=4x+16#

Explanation:

Given-
focus #(-3,0)#
Directrix #x+5=0#

Look at the diagram
enter image source here

General form of the equation

#y^2=4ax#

Since vertex is away from the origin

#(y-k)^2=4a(x-h)#

Where (h,k) is vertex
From the given information

#h=-4#
#k=0#
#a=1#

Substitute these values

#(y-0)^2=4xx1xx(x-(-4))#
#y^2=4(x+4)#
#y^2=4x+16#

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