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Question Number 63618 by kaivan.ahmadi last updated on 06/Jul/19

Commented by kaivan.ahmadi last updated on 06/Jul/19

i dont think if F′P−FP=0<a then P is not inside

$${i}\:{dont}\:{think}\: \\ $$$${if}\:{F}'{P}−{FP}=\mathrm{0}<{a}\:{then}\:{P}\:{is}\:{not}\:{inside} \\ $$

Commented by kaivan.ahmadi last updated on 06/Jul/19

explain the condition that the point P is inside the hyperbola.

$${explain}\:{the}\:{condition}\:{that}\:{the}\:{point}\:{P} \\ $$$${is}\:{inside}\:{the}\:{hyperbola}. \\ $$

Commented by kaivan.ahmadi last updated on 06/Jul/19

can we plot a tangent from the point P to hyperbola?

$${can}\:{we}\:{plot}\:{a}\:{tangent}\:{from}\:{the}\:{point}\:{P} \\ $$$${to}\:{hyperbola}? \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

If B be any point on hyperbola then F′B∽FB =constant (say a ) ∴ then F′P − FP < a I think.

$$\mathrm{If}\:\mathrm{B}\:\mathrm{be}\:\mathrm{any}\:\mathrm{point}\:\mathrm{on}\:\mathrm{hyperbola} \\ $$$$\mathrm{then}\:\mathrm{F}'\mathrm{B}\backsim\mathrm{FB}\:=\mathrm{constant}\:\left(\mathrm{say}\:\boldsymbol{\mathrm{a}}\:\right) \\ $$$$\therefore\:\mathrm{then}\:\mathrm{F}'\mathrm{P}\:−\:\mathrm{FP}\:<\:\mathrm{a} \\ $$$$\mathrm{I}\:\mathrm{think}. \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

Try F′P−FP> a

$$\mathrm{Try} \\ $$$$\mathrm{F}'\mathrm{P}−\mathrm{FP}>\:\mathrm{a} \\ $$

Answered by MJS last updated on 06/Jul/19

hyperbola a, b, e>0 y=±(b/a)(√(x^2 −a^2 )) ∣F_1 F_2 ∣=2e H∈hyp: ∣F_1 H∣−∣F_2 H∣=2a a^2 +b^2 =e^2 if we fix e ⇒ b=(√(e^2 −a^2 )); b>0 ⇒ a<e y=±((√((e^2 −a^2 )(x^2 −a^2 )))/a) for lower values of a the hyperbola becomes steeper (or more open) ⇒ for P on a hyperbola inside the given one we would need a higher value of a ⇒ ∣F_1 P∣−∣F_2 P∣>2a

$$\mathrm{hyperbola} \\ $$$${a},\:{b},\:{e}>\mathrm{0} \\ $$$${y}=\pm\frac{{b}}{{a}}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\mid{F}_{\mathrm{1}} {F}_{\mathrm{2}} \mid=\mathrm{2}{e} \\ $$$${H}\in\mathrm{hyp}:\:\mid{F}_{\mathrm{1}} {H}\mid−\mid{F}_{\mathrm{2}} {H}\mid=\mathrm{2}{a} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={e}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{fix}\:{e}\:\Rightarrow\:{b}=\sqrt{{e}^{\mathrm{2}} −{a}^{\mathrm{2}} };\:{b}>\mathrm{0}\:\Rightarrow\:{a}<{e} \\ $$$${y}=\pm\frac{\sqrt{\left({e}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}}{{a}} \\ $$$$\mathrm{for}\:\mathrm{lower}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{the}\:\mathrm{hyperbola}\:\mathrm{becomes} \\ $$$$\mathrm{steeper}\:\left(\mathrm{or}\:\mathrm{more}\:\mathrm{open}\right) \\ $$$$\Rightarrow\:\mathrm{for}\:{P}\:\mathrm{on}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{given}\:\mathrm{one} \\ $$$$\mathrm{we}\:\mathrm{would}\:\mathrm{need}\:\mathrm{a}\:\mathrm{higher}\:\mathrm{value}\:\mathrm{of}\:{a} \\ $$$$\Rightarrow\:\mid{F}_{\mathrm{1}} {P}\mid−\mid{F}_{\mathrm{2}} {P}\mid>\mathrm{2}{a} \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

great sir . Sir please suggest something on the solution of the problem of the comment of63574

$$\mathrm{great}\:\mathrm{sir}\:.\:\mathrm{Sir}\:\mathrm{please}\:\mathrm{suggest}\:\mathrm{something}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{of}\:\mathrm{the}\:\mathrm{comment}\:\mathrm{of63574} \\ $$

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