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Question Number 51492 by peter frank last updated on 27/Dec/18

For ellipse 16x^2 +4y^2 +96x−8y−84=0 find i)centre ii)verteces iii)focus iv)directrix v)length of major and minor axis vi)ecentricity vii)graph the ellipse

$${For}\:{ellipse}\: \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{96}{x}−\mathrm{8}{y}−\mathrm{84}=\mathrm{0} \\ $$$${find} \\ $$$$\left.{i}\right){centre} \\ $$$$\left.{ii}\right){verteces} \\ $$$$\left.{iii}\right){focus} \\ $$$$\left.{iv}\right){directrix} \\ $$$$\left.{v}\right){length}\:{of}\:{major}\: \\ $$$${and}\:{minor}\:{axis} \\ $$$$\left.{vi}\right){ecentricity} \\ $$$$\left.{vii}\right){graph}\:{the}\:{ellipse} \\ $$

Answered by peter frank last updated on 27/Dec/18

16x^2 +96x+4y^2 −8y−84=0 16(x^2 +6x+9)+4(y^2 −2y+1)=84+144+4 16(x+3)^2 +4(y−1)^2 =232 (((x+3)^2 )/((232)/(16)))+(((y−1)^2 )/((232)/4))=1 a=((√(232))/4) b=((√(232))/2) a^2 =b^2 (1−e^2 ) e=(3/4) (((x−p))/a^2 )+(((y−q))/b^2 )=0 centre=(p,q)=(−3,1) verteces=(±a,0)+(p,q) verteces=±(0.808,1) focus=±(ae,0)+(p,q) focus=±(0.264,1) derectrix x−p=±(a/e)=3±((√(232))/3) length of major=2a=((√(232))/2) unit minor axis=2b=(√(232)) unit

$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{96}{x}+\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}−\mathrm{84}=\mathrm{0} \\ $$$$\mathrm{16}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right)=\mathrm{84}+\mathrm{144}+\mathrm{4} \\ $$$$\mathrm{16}\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{232} \\ $$$$\frac{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }{\frac{\mathrm{232}}{\mathrm{16}}}+\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\frac{\mathrm{232}}{\mathrm{4}}}=\mathrm{1} \\ $$$${a}=\frac{\sqrt{\mathrm{232}}}{\mathrm{4}}\:\:\:\:\:{b}=\frac{\sqrt{\mathrm{232}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right) \\ $$$${e}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\left({x}−{p}\right)}{{a}^{\mathrm{2}} }+\frac{\left({y}−{q}\right)}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$${centre}=\left({p},{q}\right)=\left(−\mathrm{3},\mathrm{1}\right) \\ $$$${verteces}=\left(\pm{a},\mathrm{0}\right)+\left({p},{q}\right) \\ $$$${verteces}=\pm\left(\mathrm{0}.\mathrm{808},\mathrm{1}\right) \\ $$$${focus}=\pm\left({ae},\mathrm{0}\right)+\left({p},{q}\right) \\ $$$${focus}=\pm\left(\mathrm{0}.\mathrm{264},\mathrm{1}\right) \\ $$$${derectrix} \\ $$$$\:{x}−{p}=\pm\frac{{a}}{{e}}=\mathrm{3}\pm\frac{\sqrt{\mathrm{232}}}{\mathrm{3}} \\ $$$${length}\:{of}\:{major}=\mathrm{2}{a}=\frac{\sqrt{\mathrm{232}}}{\mathrm{2}}\:{unit} \\ $$$$\:{minor}\:{axis}=\mathrm{2}{b}=\sqrt{\mathrm{232}}\:{unit} \\ $$$$ \\ $$

Commented by ajfour last updated on 27/Dec/18

great, u′ve solved yourselves!

$${great},\:{u}'{ve}\:{solved}\:{yourselves}! \\ $$

Commented by peter frank last updated on 28/Dec/18

yes sir...please check right wrong?

$${yes}\:{sir}...{please}\:{check}\:{right} \\ $$$${wrong}? \\ $$

Commented by peter frank last updated on 28/Dec/18

mrW and ajfour help to graph the ellipse

$${mrW}\:{and}\:{ajfour}\:{help}\:{to} \\ $$$${graph}\:{the}\:{ellipse} \\ $$

Commented by mr W last updated on 28/Dec/18

correct sir!

$${correct}\:{sir}! \\ $$

Commented by peter frank last updated on 28/Dec/18

thanks

$${thanks} \\ $$

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