Google
Câu hỏi: Cho hypebol \((H): \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\). Gọi \(F_1, F_2\) là các tiêu điểm và \(A_1, A_2\) là các đỉnh của \((H)\). \(M\) là điểm tùy ý trên \((H)\) có hình chiếu trên \(Ox\) là \(N\). Chứng minh rằng
a) \(O{M^2} - M{F_1}. M{F_2} = {a^2} - {b^2}\);
b) \({(M{F_1} + M{F_2})^2} = 4(O{M^2} + {b^2})\);
c) \(N{M^2} = \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}} \).
a) \(O{M^2} - M{F_1}. M{F_2} = {a^2} - {b^2}\);
b) \({(M{F_1} + M{F_2})^2} = 4(O{M^2} + {b^2})\);
c) \(N{M^2} = \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}} \).
Lời giải chi tiết
(1.116).
\(M(x; y) \in (H) \Leftrightarrow \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1, \)
\( M{F_1} = \left| {a + \dfrac{c}{a}x} \right| , M{F_2} = \left| {a - \dfrac{c}{a}x} \right|.\)
a)Ta có
\(\begin{array}{l}O{M^2} - M{F_1}. M{F_2}\\ = {x^2} + {y^2} - \left| {{a^2} - \dfrac{{{c^2}}}{{{a^2}}}{x^2}} \right|\\ = {x^2} + {y^2} - \left| {{a^2} - {c^2}\left( {1 + \dfrac{{{y^2}}}{{{b^2}}}} \right)} \right|\\ = {x^2} + {y^2} - \left| { - {b^2} - \dfrac{{{c^2}}}{{{b^2}}}{y^2}} \right|\\ = {x^2} + {y^2} - {b^2} - \dfrac{{{c^2}}}{{{b^2}}}{y^2}\\ = {a^2} + \dfrac{{{a^2}}}{{{b^2}}}{y^2} + {y^2} - {b^2} - \dfrac{{{a^2} + {b^2}}}{{{b^2}}}{y^2}\\ = {a^2} - {b^2}.\end{array}\)
b)
\(\begin{array}{l}{(M{F_1} + M{F_2})^2} \\= {(M{F_1} - M{F_2})^2} + 4M{F_1}. M{F_2}\\ = 4{a^2} + 4\left| {{a^2} - \dfrac{{{c^2}}}{{{a^2}}}{x^2}} \right|\\= 4{a^2} + 4{b^2} + \dfrac{{4{c^2}}}{{{b^2}}}{y^2} (1)\\4(O{M^2} + {b^2}) = 4({x^2} + {y^2} + {b^2}) \\= 4{x^2} + 4{y^2} + 4{b^2}\\= 4\left({{a^2} + \dfrac{{{a^2}}}{{{b^2}}}{y^2}} \right) + 4{y^2} + 4{b^2}\\= 4{a^2} + 4{b^2} + 4{y^2}\left({ \dfrac{{{a^2}}}{{{b^2}}} + 1} \right)\\= 4{a^2} + 4{b^2} + \dfrac{{4{c^2}}}{{{b^2}}}{y^2}.(2)\end{array}\)
Từ (1) và (2) suy ra điều phải chứng minh.
c)
\(\begin{array}{l}M{N^2} = {y^2}.\\ \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}}\\ = \dfrac{{{b^2}}}{{{a^2}}}( - x - a)(- x + a)\\ = - \dfrac{{{b^2}}}{{{a^2}}}({a^2} - {x^2}) = - {b^2} + \dfrac{{{b^2}}}{{{a^2}}}{x^2}\\= - {b^2} + {b^2}\left({1 + \dfrac{{{y^2}}}{{{b^2}}}} \right) = {y^2}.\end{array}\)
Vậy \(N{M^2} = \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}} \).
(1.116).
\(M(x; y) \in (H) \Leftrightarrow \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1, \)
\( M{F_1} = \left| {a + \dfrac{c}{a}x} \right| , M{F_2} = \left| {a - \dfrac{c}{a}x} \right|.\)
a)Ta có
\(\begin{array}{l}O{M^2} - M{F_1}. M{F_2}\\ = {x^2} + {y^2} - \left| {{a^2} - \dfrac{{{c^2}}}{{{a^2}}}{x^2}} \right|\\ = {x^2} + {y^2} - \left| {{a^2} - {c^2}\left( {1 + \dfrac{{{y^2}}}{{{b^2}}}} \right)} \right|\\ = {x^2} + {y^2} - \left| { - {b^2} - \dfrac{{{c^2}}}{{{b^2}}}{y^2}} \right|\\ = {x^2} + {y^2} - {b^2} - \dfrac{{{c^2}}}{{{b^2}}}{y^2}\\ = {a^2} + \dfrac{{{a^2}}}{{{b^2}}}{y^2} + {y^2} - {b^2} - \dfrac{{{a^2} + {b^2}}}{{{b^2}}}{y^2}\\ = {a^2} - {b^2}.\end{array}\)
b)
\(\begin{array}{l}{(M{F_1} + M{F_2})^2} \\= {(M{F_1} - M{F_2})^2} + 4M{F_1}. M{F_2}\\ = 4{a^2} + 4\left| {{a^2} - \dfrac{{{c^2}}}{{{a^2}}}{x^2}} \right|\\= 4{a^2} + 4{b^2} + \dfrac{{4{c^2}}}{{{b^2}}}{y^2} (1)\\4(O{M^2} + {b^2}) = 4({x^2} + {y^2} + {b^2}) \\= 4{x^2} + 4{y^2} + 4{b^2}\\= 4\left({{a^2} + \dfrac{{{a^2}}}{{{b^2}}}{y^2}} \right) + 4{y^2} + 4{b^2}\\= 4{a^2} + 4{b^2} + 4{y^2}\left({ \dfrac{{{a^2}}}{{{b^2}}} + 1} \right)\\= 4{a^2} + 4{b^2} + \dfrac{{4{c^2}}}{{{b^2}}}{y^2}.(2)\end{array}\)
Từ (1) và (2) suy ra điều phải chứng minh.
c)
\(\begin{array}{l}M{N^2} = {y^2}.\\ \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}}\\ = \dfrac{{{b^2}}}{{{a^2}}}( - x - a)(- x + a)\\ = - \dfrac{{{b^2}}}{{{a^2}}}({a^2} - {x^2}) = - {b^2} + \dfrac{{{b^2}}}{{{a^2}}}{x^2}\\= - {b^2} + {b^2}\left({1 + \dfrac{{{y^2}}}{{{b^2}}}} \right) = {y^2}.\end{array}\)
Vậy \(N{M^2} = \dfrac{{{b^2}}}{{{a^2}}}.\overline {N{A_1}} .\overline {N{A_2}} \).