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Category: Coordinate Geometry

Question-57958

Question Number 57958 by rahul 19 last updated on 15/Apr/19 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19 $$\left(\mathrm{1},\mathrm{4}\right)\:\:{and}\left(\mathrm{3},\mathrm{8}\right) \\ $$$${chord}\:{eqn}\:\:{y}−\mathrm{4}=\left(\frac{\mathrm{8}−\mathrm{4}}{\mathrm{3}−\mathrm{1}}\right)\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{4}=\mathrm{2}{x}−\mathrm{2} \\ $$$${y}=\mathrm{2}{x}+\mathrm{2} \\…

Question-123040

Question Number 123040 by ajfour last updated on 22/Nov/20 Commented by ajfour last updated on 22/Nov/20 $${Find}\:{maximum}\:{side}\:{length}\:{of} \\ $$$${equilateral}\:\bigtriangleup{ABC}\:,\:{in}\:{terms}\:{of} \\ $$$${p},\:{q},\:{and}\:{r};\:{the}\:{vertices}\:{of} \\ $$$${which}\:{lie}\:{respectively}\:{on}\:{three} \\ $$$${concentric}\:{circles}\:{of}\:{radii}\:…

Tangents-are-drawn-to-x-2-y-2-16-from-the-point-P-0-h-These-tangents-meet-the-x-axis-at-A-and-B-If-area-of-PAB-is-minimum-then-find-value-of-h-

Question Number 57289 by rahul 19 last updated on 01/Apr/19 $${Tangents}\:{are}\:{drawn}\:{to}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{16}\:{from} \\ $$$${the}\:{point}\:{P}\left(\mathrm{0},{h}\right).{These}\:{tangents}\:{meet} \\ $$$${the}\:{x}−{axis}\:{at}\:{A}\:{and}\:{B}.\:{If}\:{area}\:{of}\:\Delta{PAB} \\ $$$${is}\:{minimum}\:{then}\:{find}\:{value}\:{of}\:{h}\:? \\ $$ Commented by mr W…

Question-57212

Question Number 57212 by Tinkutara last updated on 31/Mar/19 Answered by ajfour last updated on 31/Mar/19 $$\mathrm{a}−\mathrm{2}\sqrt{\mathrm{bc}}=\mathrm{b}+\mathrm{c} \\ $$$$\Rightarrow\:\:\mathrm{c}+\mathrm{2}\sqrt{\mathrm{b}}\sqrt{\mathrm{c}}+\left(\mathrm{b}−\mathrm{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\sqrt{\mathrm{c}}=\frac{−\mathrm{2}\sqrt{\mathrm{b}}\pm\sqrt{\mathrm{4b}−\mathrm{4}\left(\mathrm{b}−\mathrm{a}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\sqrt{\mathrm{c}}=−\sqrt{\mathrm{b}}\pm\sqrt{\mathrm{a}} \\ $$$$\mathrm{but}\:\mathrm{since}\:\sqrt{\mathrm{c}}\:>\mathrm{0}\:\mathrm{so}\:\mathrm{we}\:\mathrm{just}\:\mathrm{consider}…