Question Number 144201 by bramlexs22 last updated on 23/Jun/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles} \\ $$$$\mathrm{passing}\:\mathrm{through}\:\left(−\mathrm{4},\mathrm{3}\right)\:\mathrm{and}\:\mathrm{touching} \\ $$$$\mathrm{the}\:\mathrm{lines}\:\mathrm{x}+\mathrm{y}=\mathrm{2}\:\mathrm{and}\:\mathrm{x}−\mathrm{y}=\mathrm{2} \\ $$ Answered by benjo_mathlover last updated on 23/Jun/21 $$\mathrm{let}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{be}\: \\…
Question Number 131420 by ajfour last updated on 04/Feb/21 Commented by ajfour last updated on 04/Feb/21 $${In}\:{terms}\:{of}\:{ellipse}\:{parameters} \\ $$$${a}\:{and}\:{b},\:{find}\:{radius}\:{of}\:{the} \\ $$$${equal}\:{radii}\:{circles}. \\ $$ Answered by…
Question Number 279 by amandeep last updated on 25/Jan/15 $$\mathrm{If}\:{P}\:\left({x},{y}\right),\:{F}_{\mathrm{1}} =\left(\mathrm{3},\mathrm{0}\right),\:{F}_{\mathrm{2}} =\left(−\mathrm{3},\mathrm{0}\right)\:\mathrm{and}\: \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{25}{y}^{\mathrm{2}} =\mathrm{400}\:\mathrm{then}\:{PF}_{\mathrm{1}} +{PF}_{\mathrm{2}} =? \\ $$ Answered by 123456 last updated…
Question Number 229 by ssahoo last updated on 25/Jan/15 $$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{traingle}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}\:\mid{z}\mid=\mathrm{2}\:\mathrm{and}\:\mathrm{if} \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} . \\ $$ Answered by 123456…
Question Number 91 by vkulkarni last updated on 25/Jan/15 $$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{joining}\:\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{3},−\mathrm{4}\right)\:\mathrm{and}\:\:\left(−\mathrm{5},\mathrm{6}\right) \\ $$$$\mathrm{and}\:\mathrm{has}\:\mathrm{a}\:\mathrm{slope}\:\mathrm{of}\:\frac{\mathrm{3}}{\mathrm{4}}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}. \\ $$ Answered by ssahoo last updated on 04/Dec/14 $$\mathrm{slope}\:{m}=\frac{\mathrm{3}}{\mathrm{4}} \\…
Question Number 144074 by bobhans last updated on 21/Jun/21 $$\:\mathrm{The}\:\mathrm{parallelogram}\:\mathrm{ABCD}\:\mathrm{has} \\ $$$$\mid\mid\mathrm{AB}\mid\mid\:=\mathrm{6},\:\mid\mid\mathrm{AC}\mid\mid=\mathrm{7}\:\&\:\mathrm{d}\left(\mathrm{D},\mathrm{AC}\right)=\mathrm{2} \\ $$$$\mathrm{Find}\:\mathrm{d}\left(\mathrm{D},\mathrm{AB}\right). \\ $$ Answered by liberty last updated on 21/Jun/21 $$\mathrm{Area}_{\Delta\mathrm{ADC}} \:=\:\frac{\mathrm{2}×\mathrm{7}}{\mathrm{2}}=\mathrm{7}…
Question Number 12962 by tawa last updated on 07/May/17 Answered by sandy_suhendra last updated on 09/May/17 Commented by sandy_suhendra last updated on 09/May/17 $$\left.\mathrm{a}\right)\:\mathrm{from}\:\mathrm{picture}\:\left(\mathrm{i}\right) \\…
Question Number 143993 by cherokeesay last updated on 20/Jun/21 Answered by mr W last updated on 20/Jun/21 Commented by mr W last updated on 20/Jun/21…
Question Number 78453 by mathocean1 last updated on 17/Jan/20 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{points} \\ $$$$\mathrm{A}\left(−\mathrm{5};−\mathrm{5}\right)\:\mathrm{B}\left(−\mathrm{5};\mathrm{10}\right)\:\mathrm{C}\left(\mathrm{15};−\mathrm{5}\right). \\ $$$$\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equtions}\:\mathrm{of}\:\left(\mathrm{AB}\right);\:\left(\mathrm{AC}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{BC}\right)\:\mathrm{are}\:\mathrm{respectively} \\ $$$$\mathrm{x}=−\mathrm{5} \\ $$$$\mathrm{y}=−\mathrm{5} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{5} \\ $$$$ \\…
Question Number 12902 by tawa last updated on 06/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $$\left.\mathrm{1}\right)\:\boldsymbol{{c}}=\mathrm{0}\:\:\:\:\:\:\:\left(\boldsymbol{{x}}=\mathrm{0}\Rightarrow\boldsymbol{{y}}=\mathrm{0}\right) \\ $$$$\boldsymbol{{y}}^{'} =\mathrm{2}\boldsymbol{{ax}}+\boldsymbol{{b}}\:\Downarrow \\ $$$$\left.\mathrm{2}\right)\begin{cases}{\mathrm{2}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{4}}\\{\mathrm{4}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{5}}\end{cases}\Rightarrow\mathrm{2}\boldsymbol{{a}}=\mathrm{1}\Rightarrow\boldsymbol{{a}}=\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{b}}=\mathrm{3} \\ $$$$\boldsymbol{{y}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\:\:\:.\:\blacksquare…