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Question Number 112664 by bemath last updated on 09/Sep/20
what the equation of parabola   whose focus F(−3,4) and directrix  is 3x−4y+5=0 ?
$$\mathrm{what}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\: \\ $$$$\mathrm{whose}\:\mathrm{focus}\:\mathrm{F}\left(−\mathrm{3},\mathrm{4}\right)\:\mathrm{and}\:\mathrm{directrix} \\ $$$$\mathrm{is}\:\mathrm{3x}−\mathrm{4y}+\mathrm{5}=\mathrm{0}\:? \\ $$
Commented by bemath last updated on 09/Sep/20
Answered by bobhans last updated on 09/Sep/20
(⧫) (√((x+3)^2 +(y−4)^2 )) = ∣((3x−4y+5)/5)∣  squaring ⇒(x−3)^2 +(y−4)^2  =(((3x−4y+5)^2 )/(25))  ⇒25(x^2 +y^2 −6x−8y+25)=9x^2 +16y^2 +25−24xy+30x−40y  ⇒16x^2 +9y^2 +24xy−150x−160y+600=0
$$\left(\blacklozenge\right)\:\sqrt{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\mid\frac{\mathrm{3x}−\mathrm{4y}+\mathrm{5}}{\mathrm{5}}\mid \\ $$$$\mathrm{squaring}\:\Rightarrow\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} \:=\frac{\left(\mathrm{3x}−\mathrm{4y}+\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{25}} \\ $$$$\Rightarrow\mathrm{25}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}−\mathrm{8y}+\mathrm{25}\right)=\mathrm{9x}^{\mathrm{2}} +\mathrm{16y}^{\mathrm{2}} +\mathrm{25}−\mathrm{24xy}+\mathrm{30x}−\mathrm{40y} \\ $$$$\Rightarrow\mathrm{16x}^{\mathrm{2}} +\mathrm{9y}^{\mathrm{2}} +\mathrm{24xy}−\mathrm{150x}−\mathrm{160y}+\mathrm{600}=\mathrm{0} \\ $$

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