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^2 +y^2 −6x+6y−16=0 ⇔(x−3)^2 +(y+3)^2 =34 center C(3,−3), radius R=(√(34)) line: 4x−y=15 ⇔y=4x−15 4×3−(−3)=15 ⇒line passes through C. intersection of circle and line: (x−3)^2 +(4x−15+3)^2 =34 17(x−3)^2 =34 x−3=±(√2) ⇒x=3±(√2) ⇒y=12±4(√2)−15=−3±4(√2) M(3−(√2),−3−4(√2)), N(3+(√2),−3+4(√2)) distance from N to line y=x (i.e. x−y=0): d_N =((3+(√2)−(−3+4(√2)))/(√(1^2 +(−1)^2 )))=((6−3(√2))/(√2))=3((√2)−1) point G with MG⊥line y=x: x_G =x_M +λ y_G =y_M −λ ⇒y_M −λ=x_M +λ ⇒λ=((y_M −x_M )/2) ⇒x_G =x_M +((y_M −x_M )/2)=((x_M +y_M )/2)=((3−(√2)−3−4(√2))/2)=−((5(√2))/2) ⇒y_G =x_G =−((5(√2))/2) Eqn. of ellipse: the ellipse is following ellipse after a rotation of (π/4): (((x−k)^2 )/a^2 )+(y^2 /b^2 )=1 where b=d_N =3((√2)−1) Eqn. after rotation of (π/4): (((((x+y)/(√2))−k)^2 )/a^2 )+(((((−x+y)/(√2)))^2 )/b^2 )=1 (((x+y−(√2) k)^2 )/a^2 )+(((−x+y)^2 )/b^2 )=2 Point G is on the ellipse, (((−5(√2)−(√2) k)^2 )/a^2 )=2 (((5+k)^2 )/a^2 )=1 ⇒k=a−5 Line y=4x−15 is tangent of ellipse: (((x+4x−15−(√2) k)^2 )/a^2 )+(((−x+4x−15)^2 )/b^2 )=2 (((5x−15−(√2) k)^2 )/a^2 )+(((3x−15)^2 )/b^2 )=2 b^2 (5x−15−(√2)k)^2 +a^2 (3x−15)^2 =2a^2 b^2 b^2 [25x^2 −10(15+(√2)k)x+(15+(√2)k)^2 ]+a^2 [9x^2 −90x+225]=2a^2 b^2 (25b^2 +9a^2 )x^2 −10[(15+(√2)k)b^2 +9a^2 ]x+[(15+(√2)k)^2 b^2 +(225−2b^2 )a^2 ]=0 for a tangent there is only one intersection point, Δ=100[(15+(√2)k)b^2 +9a^2 ]^2 −4(25b^2 +9a^2 )[(15+(√2)k)^2 b^2 +(225−2b^2 )a^2 ]=0 k=a−5 b^2 =9(3−2(√2)) ⇒25[9(3−2(√2)){15+(√2)(a−5)}+9a^2 ]^2 −[225(3−2(√2))+9a^2 ][9(3−2(√2)){15+(√2) (a−5)}^2 +(171+36(√2))a^2 ]=0 ⇒((√2)−1)a^3 −10(3(√2) −4)a^2 =0 ⇒a=((10(3(√2)−4))/((√2)−1))=10(2−(√2))≈5.86 b=3((√2)−1)≈1.24 (b/a)=((3((√2)−1))/(10(2−(√2))))=((3(√2))/(20)) eccentricity=(c/a)=((√(a^2 −b^2 ))/a)=(√(1−((b/a))^2 ))=(√(1−(9/(200))))≈0.977 length of latus rectum =((2b^2 )/a)=((18(3−2(√2)))/(10(2−(√2))))=((18−9(√2))/(10)) k=a−5=5(3−2(√2))≈0.86 (√2)k=5(3(√2)−4) therefore the eqn. of ellipse (before rotation) is (([x−5(3−2(√2))]^2 )/(200(3−2(√2))))+(y^2 /(9(3−2(√2))))=1 center of ellipse is (k,0)=(5(3−2(√2)),0) the eqn. of the searched ellipse is (([x+y−5(3(√2)−4)]^2 )/(200))+(((y−x)^2 )/9)=2(3−2(√2)) center of ellipse is ((k/(√2)),(k/(√2)))=(((5(3(√2)−4))/2),((5(3(√2)−4))/2))](Q28232.png)
$${circle}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}{y}−\mathrm{16}=\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$${center}\:{C}\left(\mathrm{3},−\mathrm{3}\right),\:{radius}\:{R}=\sqrt{\mathrm{34}} \\ $$$$ \\ $$$${line}: \\ $$$$\mathrm{4}{x}−{y}=\mathrm{15} \\ $$$$\Leftrightarrow{y}=\mathrm{4}{x}−\mathrm{15} \\ $$$$\mathrm{4}×\mathrm{3}−\left(−\mathrm{3}\right)=\mathrm{15} \\ $$$$\Rightarrow{line}\:{passes}\:{through}\:{C}. \\ $$$$ \\ $$$${intersection}\:{of}\:{circle}\:{and}\:{line}: \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{4}{x}−\mathrm{15}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$$\mathrm{17}\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$${x}−\mathrm{3}=\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{3}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{12}\pm\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{15}=−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${M}\left(\mathrm{3}−\sqrt{\mathrm{2}},−\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}}\right),\:{N}\left(\mathrm{3}+\sqrt{\mathrm{2}},−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$${distance}\:{from}\:{N}\:{to}\:{line}\:{y}={x}\:\left({i}.{e}.\:{x}−{y}=\mathrm{0}\right): \\ $$$${d}_{{N}} =\frac{\mathrm{3}+\sqrt{\mathrm{2}}−\left(−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}}=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$ \\ $$$${point}\:{G}\:{with}\:{MG}\bot{line}\:{y}={x}: \\ $$$${x}_{{G}} ={x}_{{M}} +\lambda \\ $$$${y}_{{G}} ={y}_{{M}} −\lambda \\ $$$$\Rightarrow{y}_{{M}} −\lambda={x}_{{M}} +\lambda \\ $$$$\Rightarrow\lambda=\frac{{y}_{{M}} −{x}_{{M}} }{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{G}} ={x}_{{M}} +\frac{{y}_{{M}} −{x}_{{M}} }{\mathrm{2}}=\frac{{x}_{{M}} +{y}_{{M}} }{\mathrm{2}}=\frac{\mathrm{3}−\sqrt{\mathrm{2}}−\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}=−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{G}} ={x}_{{G}} =−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{ellipse}: \\ $$$${the}\:{ellipse}\:{is}\:{following}\:{ellipse}\:{after}\: \\ $$$${a}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{4}}: \\ $$$$\frac{\left({x}−{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${where}\:{b}={d}_{{N}} =\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${Eqn}.\:{after}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{4}}: \\ $$$$\frac{\left(\frac{{x}+{y}}{\sqrt{\mathrm{2}}}−{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\frac{−{x}+{y}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\left({x}+{y}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(−{x}+{y}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$$ \\ $$$${Point}\:{G}\:{is}\:{on}\:{the}\:{ellipse}, \\ $$$$\frac{\left(−\mathrm{5}\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\frac{\left(\mathrm{5}+{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{k}={a}−\mathrm{5} \\ $$$$ \\ $$$${Line}\:{y}=\mathrm{4}{x}−\mathrm{15}\:{is}\:{tangent}\:{of}\:{ellipse}: \\ $$$$\frac{\left({x}+\mathrm{4}{x}−\mathrm{15}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(−{x}+\mathrm{4}{x}−\mathrm{15}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\frac{\left(\mathrm{5}{x}−\mathrm{15}−\sqrt{\mathrm{2}}\:{k}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\mathrm{3}{x}−\mathrm{15}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2} \\ $$$${b}^{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{15}−\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{3}{x}−\mathrm{15}\right)^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \left[\mathrm{25}{x}^{\mathrm{2}} −\mathrm{10}\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){x}+\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} \right]+{a}^{\mathrm{2}} \left[\mathrm{9}{x}^{\mathrm{2}} −\mathrm{90}{x}+\mathrm{225}\right]=\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left(\mathrm{25}{b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{10}\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right]{x}+\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} {b}^{\mathrm{2}} +\left(\mathrm{225}−\mathrm{2}{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$ \\ $$$${for}\:{a}\:{tangent}\:{there}\:{is}\:{only}\:{one}\:{intersection}\:{point}, \\ $$$$\Delta=\mathrm{100}\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right){b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right]^{\mathrm{2}} −\mathrm{4}\left(\mathrm{25}{b}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \right)\left[\left(\mathrm{15}+\sqrt{\mathrm{2}}{k}\right)^{\mathrm{2}} {b}^{\mathrm{2}} +\left(\mathrm{225}−\mathrm{2}{b}^{\mathrm{2}} \right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$${k}={a}−\mathrm{5} \\ $$$${b}^{\mathrm{2}} =\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{25}\left[\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left\{\mathrm{15}+\sqrt{\mathrm{2}}\left({a}−\mathrm{5}\right)\right\}+\mathrm{9}{a}^{\mathrm{2}} \right]^{\mathrm{2}} −\left[\mathrm{225}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)+\mathrm{9}{a}^{\mathrm{2}} \right]\left[\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left\{\mathrm{15}+\sqrt{\mathrm{2}}\:\left({a}−\mathrm{5}\right)\right\}^{\mathrm{2}} +\left(\mathrm{171}+\mathrm{36}\sqrt{\mathrm{2}}\right){a}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){a}^{\mathrm{3}} −\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}\:−\mathrm{4}\right){a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\approx\mathrm{5}.\mathrm{86} \\ $$$${b}=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\approx\mathrm{1}.\mathrm{24} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{20}} \\ $$$${eccentricity}=\frac{{c}}{{a}}=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{a}}=\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{200}}}\approx\mathrm{0}.\mathrm{977} \\ $$$${length}\:{of}\:{latus}\:{rectum}\:=\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}=\frac{\mathrm{18}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{18}−\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{10}} \\ $$$$ \\ $$$${k}={a}−\mathrm{5}=\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\approx\mathrm{0}.\mathrm{86} \\ $$$$\sqrt{\mathrm{2}}{k}=\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right) \\ $$$${therefore}\:{the}\:{eqn}.\:{of}\:{ellipse}\:\left({before}\right. \\ $$$$\left.{rotation}\right)\:{is} \\ $$$$\frac{\left[{x}−\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\right]^{\mathrm{2}} }{\mathrm{200}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}=\mathrm{1} \\ $$$${center}\:{of}\:{ellipse}\:{is}\:\left({k},\mathrm{0}\right)=\left(\mathrm{5}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right),\mathrm{0}\right) \\ $$$$ \\ $$$${the}\:{eqn}.\:{of}\:{the}\:{searched}\:{ellipse}\:{is} \\ $$$$\frac{\left[{x}+{y}−\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\right]^{\mathrm{2}} }{\mathrm{200}}+\frac{\left({y}−{x}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${center}\:{of}\:{ellipse}\:{is}\:\left(\frac{{k}}{\sqrt{\mathrm{2}}},\frac{{k}}{\sqrt{\mathrm{2}}}\right)=\left(\frac{\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{2}},\frac{\mathrm{5}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{2}}\right) \\ $$
Commented by mrW2 last updated on 22/Jan/18
