What is the equation of the parabola with a focus at (8,2) and a directrix of y= 5?

2 Answers
Narad T.
Oct 26, 2017

The equation is #(x-8)^2=-3(2y-7)#

Explanation:

Any point on the parabola is equidistant from the focus and the directrix

Therefore,

#sqrt((x-8)+(y-2))=5-y#

Squaring,

#(x-8)^2+(y-2)^2=(5-y)^2#

#(x-8)^2+cancely^2-4y+4=25-10y+cancely^2#

#(x-8)^2=-6y+21#

#(x-8)^2=-3(2y-7)#

graph{((x-8)^2+3(2y-7))(y-5)((x-8)^2+(y-2)^2-0.1)=0 [-32.47, 32.47, -16.24, 16.25]}

Jim G.
Oct 26, 2017

#x^2-16x+6y+43=0#

Explanation:

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

#"the distance from "(x,y)" to the focus and directrix"#
#"are equal"#

#"using the "color(blue)"distance formula"" and equating"#

#rArrsqrt((x-8)^2+(y-2)^2)=|y-5|#

#color(blue)"squaring both sides"#

#(x-8)^2+(y-2)^2=(y-5)^2#

#rArrx^2-16x+64+y^2-4y+4=y^2-10y+25#

#rArrx^2-16x+64cancel(+y^2)-4y+4cancel(-y^2)+10y-25=0#

#rArrx^2-16x+6y+43=0#

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
4544 views around the world
You can reuse this answer
Creative Commons License