Question Number 209125 by Spillover last updated on 02/Jul/24 | ||
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Answered by mr W last updated on 03/Jul/24 | ||
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Commented by mr W last updated on 05/Jul/24 | ||
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$${AB}={AF}={a} \\ $$$${a}+{a}\:\mathrm{cos}\:\theta=\mathrm{2}{p} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{2}{p}}{{a}}−\mathrm{1} \\ $$ | ||
Commented by Spillover last updated on 03/Jul/24 | ||
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$${nice}\:{solution}\:{thank}\:{you} \\ $$ | ||
Commented by Spillover last updated on 03/Jul/24 | ||
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Answered by Spillover last updated on 03/Jul/24 | ||
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Commented by Spillover last updated on 04/Jul/24 | ||
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$${thank}\:{you}\:{very}\:{for}\:{your}\:{clarification}. \\ $$ | ||
Commented by mr W last updated on 03/Jul/24 | ||
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$${but}\:\mathrm{cos}\:\theta=−\mathrm{1}+\frac{\mathrm{2}{p}}{{a}} \\ $$ | ||
Commented by Spillover last updated on 03/Jul/24 | ||
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$${why}\:\mathrm{2}{p}?\:{what}\:{is}\:{reason}\:{behind} \\ $$ | ||
Commented by mr W last updated on 03/Jul/24 | ||
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$${distance}\:{from}\:{focus}\:{to}\:{vertex}\:{is}\:{p} \\ $$$${distance}\:{from}\:{focus}\:{to}\:{directrix}\:{is}\:\mathrm{2}{p} \\ $$$${see}\:{also}\:{my}\:{diagram}\:{above}. \\ $$ | ||
Commented by mr W last updated on 03/Jul/24 | ||
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Answered by Spillover last updated on 05/Jul/24 | ||
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