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Focus-and-vertex-of-a-parabola-are-at-3-4-and-0-0-Find-the-equation-of-the-directrix-

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Question Number 203465 by princemurtuja last updated on 19/Jan/24
Focus and vertex of a parabola are at (3, 4) and (0,0). Find the equation of the directrix.
$$\mathrm{Focus}\:\mathrm{and}\:\mathrm{vertex}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{are}\:\mathrm{at}\:\left(\mathrm{3},\:\mathrm{4}\right)\:\mathrm{and}\:\left(\mathrm{0},\mathrm{0}\right). \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{directrix}. \\ $$
Answered by mr W last updated on 20/Jan/24
F(3, 4) V(0, 0) directrix is ⊥ to FV. equation for directrix: 3x+4y+c=0 equstion for parabola: (x−3)^2 +(y−4)^2 =(((3x+4y+c)^2 )/(25)) (0−3)^2 +(0−4)^2 =(((3×0+4×0+c)^2 )/(25)) ⇒c=25 (−25 rejected) directrix: 3x+4y+25=0 parabola: (x−3)^2 +(y−4)^2 =(((3x+4y+25)^2 )/(25))
$${F}\left(\mathrm{3},\:\mathrm{4}\right) \\ $$$${V}\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$${directrix}\:{is}\:\bot\:{to}\:{FV}. \\ $$$${equation}\:{for}\:{directrix}: \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}+{c}=\mathrm{0} \\ $$$${equstion}\:{for}\:{parabola}: \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\frac{\left(\mathrm{3}{x}+\mathrm{4}{y}+{c}\right)^{\mathrm{2}} }{\mathrm{25}} \\ $$$$\left(\mathrm{0}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{0}−\mathrm{4}\right)^{\mathrm{2}} =\frac{\left(\mathrm{3}×\mathrm{0}+\mathrm{4}×\mathrm{0}+{c}\right)^{\mathrm{2}} }{\mathrm{25}} \\ $$$$\Rightarrow{c}=\mathrm{25}\:\:\left(−\mathrm{25}\:{rejected}\right) \\ $$$${directrix}:\: \\ $$$$\:\:\:\:\:\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{25}=\mathrm{0} \\ $$$${parabola}: \\ $$$$\:\:\:\:\:\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\frac{\left(\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{25}} \\ $$
Commented by mr W last updated on 19/Jan/24
Answered by som(math1967) last updated on 20/Jan/24
Other way intersecting point of directrix and axis⇒(−3,−4) slope of axis=((0−4)/(0−3))=(4/3) slope of directrix =((−3)/4) equation of directrix (y+4)=((−3)/4)(x+3) 4y+16=−3x−9 3x+4y+25=0
$${Other}\:{way} \\ $$$${intersecting}\:\:{point}\:{of}\:{directrix} \\ $$$${and}\:{axis}\Rightarrow\left(−\mathrm{3},−\mathrm{4}\right) \\ $$$${slope}\:{of}\:{axis}=\frac{\mathrm{0}−\mathrm{4}}{\mathrm{0}−\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${slope}\:{of}\:{directrix}\:=\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$${equation}\:{of}\:{directrix} \\ $$$$\left({y}+\mathrm{4}\right)=\frac{−\mathrm{3}}{\mathrm{4}}\left({x}+\mathrm{3}\right) \\ $$$$\mathrm{4}{y}+\mathrm{16}=−\mathrm{3}{x}−\mathrm{9} \\ $$$$\:\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{25}=\mathrm{0} \\ $$

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