Question Number 51492 by peter frank last updated on 27/Dec/18
$${For}\:{ellipse}\: \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{96}{x}−\mathrm{8}{y}−\mathrm{84}=\mathrm{0} \\ $$$${find} \\ $$$$\left.{i}\right){centre} \\ $$$$\left.{ii}\right){verteces} \\ $$$$\left.{iii}\right){focus} \\ $$$$\left.{iv}\right){directrix} \\ $$$$\left.{v}\right){length}\:{of}\:{major}\: \\ $$$${and}\:{minor}\:{axis} \\ $$$$\left.{vi}\right){ecentricity} \\ $$$$\left.{vii}\right){graph}\:{the}\:{ellipse} \\ $$
Answered by peter frank last updated on 27/Dec/18
$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{96}{x}+\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}−\mathrm{84}=\mathrm{0} \\ $$$$\mathrm{16}\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right)=\mathrm{84}+\mathrm{144}+\mathrm{4} \\ $$$$\mathrm{16}\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{232} \\ $$$$\frac{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }{\frac{\mathrm{232}}{\mathrm{16}}}+\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\frac{\mathrm{232}}{\mathrm{4}}}=\mathrm{1} \\ $$$${a}=\frac{\sqrt{\mathrm{232}}}{\mathrm{4}}\:\:\:\:\:{b}=\frac{\sqrt{\mathrm{232}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right) \\ $$$${e}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\left({x}−{p}\right)}{{a}^{\mathrm{2}} }+\frac{\left({y}−{q}\right)}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$${centre}=\left({p},{q}\right)=\left(−\mathrm{3},\mathrm{1}\right) \\ $$$${verteces}=\left(\pm{a},\mathrm{0}\right)+\left({p},{q}\right) \\ $$$${verteces}=\pm\left(\mathrm{0}.\mathrm{808},\mathrm{1}\right) \\ $$$${focus}=\pm\left({ae},\mathrm{0}\right)+\left({p},{q}\right) \\ $$$${focus}=\pm\left(\mathrm{0}.\mathrm{264},\mathrm{1}\right) \\ $$$${derectrix} \\ $$$$\:{x}−{p}=\pm\frac{{a}}{{e}}=\mathrm{3}\pm\frac{\sqrt{\mathrm{232}}}{\mathrm{3}} \\ $$$${length}\:{of}\:{major}=\mathrm{2}{a}=\frac{\sqrt{\mathrm{232}}}{\mathrm{2}}\:{unit} \\ $$$$\:{minor}\:{axis}=\mathrm{2}{b}=\sqrt{\mathrm{232}}\:{unit} \\ $$$$ \\ $$
Commented by ajfour last updated on 27/Dec/18
$${great},\:{u}'{ve}\:{solved}\:{yourselves}! \\ $$
Commented by peter frank last updated on 28/Dec/18
$${yes}\:{sir}…{please}\:{check}\:{right} \\ $$$${wrong}? \\ $$
Commented by peter frank last updated on 28/Dec/18
$${mrW}\:{and}\:{ajfour}\:{help}\:{to} \\ $$$${graph}\:{the}\:{ellipse} \\ $$
Commented by mr W last updated on 28/Dec/18
$${correct}\:{sir}! \\ $$
Commented by peter frank last updated on 28/Dec/18
$${thanks} \\ $$