Question Number 28185 by ajfour last updated on 21/Jan/18
$${M},\:{N}\:{are}\:{endpoints}\:{of}\:{a}\:{diameter} \\ $$$$\:\mathrm{4}{x}−{y}=\mathrm{15}\:\:{of}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}{y}−\mathrm{16}=\mathrm{0}\:;\:{and}\:{are} \\ $$$${also}\:{on}\:{the}\:{tangents}\:{at}\:{the}\:{end} \\ $$$${points}\:{of}\:{the}\:{major}\:{axis}\:{of}\:{an} \\ $$$${ellipse}\:{respectively},\:{such}\:{that} \\ $$$${MN}\:{is}\:{also}\:{tangent}\:{to}\:{the}\:{same} \\ $$$${ellipse}\:{at}\:{point}\:{P}. \\ $$$${If}\:{the}\:{major}\:{axis}\:{of}\:{the}\:{ellipse} \\ $$$${is}\:{along}\:{y}={x},\:{find} \\ $$$$\:\:\:{eccentricity},\:{length}\:{of}\:{latus} \\ $$$${rectum},\:{centre}\:{and}\:{equation}\:{of} \\ $$$${derectrices}. \\ $$
Answered by mrW2 last updated on 22/Jan/18
Commented by mrW2 last updated on 22/Jan/18
$${circle}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}{y}−\mathrm{16}=\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$${center}\:{C}\left(\mathrm{3},−\mathrm{3}\right),\:{radius}\:{R}=\sqrt{\mathrm{34}} \\ $$$$ \\ $$$${line}: \\ $$$$\mathrm{4}{x}−{y}=\mathrm{15} \\ $$$$\Leftrightarrow{y}=\mathrm{4}{x}−\mathrm{15} \\ $$$$\mathrm{4}×\mathrm{3}−\left(−\mathrm{3}\right)=\mathrm{15} \\ $$$$\Rightarrow{line}\:{passes}\:{through}\:{C}. \\ $$$$ \\ $$$${intersection}\:{of}\:{circle}\:{and}\:{line}: \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{4}{x}−\mathrm{15}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$$\mathrm{17}\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$${x}−\mathrm{3}=\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{3}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{12}\pm\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{15}=−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${M}\left(\mathrm{3}−\sqrt{\mathrm{2}},−\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}}\right),\:{N}\left(\mathrm{3}+\sqrt{\mathrm{2}},−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$${distance}\:{from}\:{N}\:{to}\:{line}\:{y}={x}\:\left({i}.{e}.\:{x}−{y}=\mathrm{0}\right): \\ $$$${d}_{{N}} =\frac{\mathrm{3}+\sqrt{\mathrm{2}}−\left(−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$ \\ $$$${point}\:{G}\:{with}\:{MG}\bot{line}\:{y}={x}: \\ $$$${x}_{{G}} ={x}_{{M}} +\lambda \\ $$$${y}_{{G}} ={y}_{{M}} −\lambda \\ $$$$\Rightarrow{y}_{{M}} −\lambda={x}_{{M}} +\lambda \\ $$$$\Rightarrow\lambda=\frac{{y}_{{M}} −{x}_{{M}} }{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{G}} ={x}_{{M}} +\frac{{y}_{{M}} −{x}_{{M}} }{\mathrm{2}}=\frac{{x}_{{M}} +{y}_{{M}} }{\mathrm{2}}=\frac{\mathrm{3}−\sqrt{\mathrm{2}}−\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}=−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{G}} ={x}_{{G}} =−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{ellipse}: \\ $$$${the}\:{ellipse}\:{is}\:{following}\:{ellipse}\:{after}\: \\ $$$${a}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{4}}: \\ $$$$\frac{\left({x}−{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${where}\:{b}={d}_{{N}} =\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${Eqn}.\:{after}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{4}}: \\ $$$$\frac{\left(\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}}−{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\frac{−{x}+{y}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\left({x}+{y}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(−{x}+{y}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$$ \\ $$$${Point}\:{G}\:{is}\:{on}\:{the}\:{ellipse}, \\ $$$$\frac{\left(−\mathrm{5}\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\frac{\left(\mathrm{5}+{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{k}={a}−\mathrm{5} \\ $$$$ \\ $$$${Line}\:{y}=\mathrm{4}{x}−\mathrm{15}\:{is}\:{tangent}\:{of}\:{ellipse}: \\ $$$$\frac{\left({x}+\mathrm{4}{x}−\mathrm{15}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(−{x}+\mathrm{4}{x}−\mathrm{15}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\frac{\left(\mathrm{5}{x}−\mathrm{15}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\mathrm{3}{x}−\mathrm{15}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$${b}^{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{15}−\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{3}{x}−\mathrm{15}\right)^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \left[\mathrm{25}{x}^{\mathrm{2}} −\mathrm{10}\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){x}+\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} \right]+{a}^{\mathrm{2}} \left[\mathrm{9}{x}^{\mathrm{2}} −\mathrm{90}{x}+\mathrm{225}\right]=\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left(\mathrm{25}{b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{10}\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right]{x}+\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} {b}^{\mathrm{2}} +\left(\mathrm{225}−\mathrm{2}{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$ \\ $$$${for}\:{a}\:{tangent}\:{there}\:{is}\:{only}\:{one}\:{intersection}\:{point}, \\ $$$$\Delta=\mathrm{100}\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right]^{\mathrm{2}} −\mathrm{4}\left(\mathrm{25}{b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right)\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} {b}^{\mathrm{2}} +\left(\mathrm{225}−\mathrm{2}{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$${k}={a}−\mathrm{5} \\ $$$${b}^{\mathrm{2}} =\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{25}\left[\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left\{\mathrm{15}+\sqrt{\mathrm{2}}\left({a}−\mathrm{5}\right)\right\}+\mathrm{9}{a}^{\mathrm{2}} \right]^{\mathrm{2}} −\left[\mathrm{225}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)+\mathrm{9}{a}^{\mathrm{2}} \right]\left[\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left\{\mathrm{15}+\sqrt{\mathrm{2}}\:\left({a}−\mathrm{5}\right)\right\}^{\mathrm{2}} +\left(\mathrm{171}+\mathrm{36}\sqrt{\mathrm{2}}\right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){a}^{\mathrm{3}} −\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}\:−\mathrm{4}\right){a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\approx\mathrm{5}.\mathrm{86} \\ $$$${b}=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\approx\mathrm{1}.\mathrm{24} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{20}} \\ $$$${eccentricity}=\frac{{c}}{{a}}=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{a}}=\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{200}}}\approx\mathrm{0}.\mathrm{977} \\ $$$${length}\:{of}\:{latus}\:{rectum}\:=\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}=\frac{\mathrm{18}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{18}−\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{10}} \\ $$$$ \\ $$$${k}={a}−\mathrm{5}=\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\approx\mathrm{0}.\mathrm{86} \\ $$$$\sqrt{\mathrm{2}}{k}=\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right) \\ $$$${therefore}\:{the}\:{eqn}.\:{of}\:{ellipse}\:\left({before}\right. \\ $$$$\left.{rotation}\right)\:{is} \\ $$$$\frac{\left[{x}−\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\right]^{\mathrm{2}} }{\mathrm{200}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}=\mathrm{1} \\ $$$${center}\:{of}\:{ellipse}\:{is}\:\left({k},\mathrm{0}\right)=\left(\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right),\mathrm{0}\right) \\ $$$$ \\ $$$${the}\:{eqn}.\:{of}\:{the}\:{searched}\:{ellipse}\:{is} \\ $$$$\frac{\left[{x}+{y}−\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\right]^{\mathrm{2}} }{\mathrm{200}}+\frac{\left({y}−{x}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${center}\:{of}\:{ellipse}\:{is}\:\left(\frac{{k}}{\:\sqrt{\mathrm{2}}},\frac{{k}}{\:\sqrt{\mathrm{2}}}\right)=\left(\frac{\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{2}},\frac{\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{2}}\right) \\ $$
Commented by mrW2 last updated on 22/Jan/18