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Contract 5
In the plane, consider the hyperbola (H) as the set of points M such that \(\left| {M{F_1} – M{F_2}} \right| = 2a\), where \({F_1}{ F_2} = 2c\) with \(c > a > 0\). We choose the coordinate system \(Oxy\) whose origin is the midpoint of the line segment \({F_1}{F_2}\). The axis \(Oy\) is the perpendicular bisector of \({F_1}{F_2}\) and \({F_2}\) lying on the ray \(Ox\) (Figure 16). Then \({F_1}( – c;0),{F_2}(c;0)\) are the focal points of the hyperbola (H)
Assuming the point \(M\left( {x;y} \right)\) belongs to the hyperbola (H), prove:
a) \(M{F_1}^2 = {x^2} + 2cx + {c^2} + {y^2}\)
b) \(M{F_2}^2 = {x^2} – 2cx + {c^2} + {y^2}\)
c) \(M{F_1}^2 – M{F_2}^2 = 4cx\)
Detailed explanation:
a) We have: \(\overrightarrow {M{F_1}} = \left( { – c – x; – y} \right) \Rightarrow M{F_1}^2 = {\left( { – c – x} \right)^2} + {y^2} = {x^2} + 2cx + {c^2} + {y^2}\)
b) We have: \(\overrightarrow {M{F_2}} = \left( {c – x; – y} \right) \Rightarrow M{F_2}^2 = {\left( {c – x} \right )^2} + {y^2} = {x^2} – 2cx + {c^2} + {y^2}\)
c) \(M{F_1}^2 – M{F_2}^2 = \left( {{x^2} + 2cx + {c^2} + {y^2}} \right) – \left( { {x^2} – 2cx + {c^2} + {y^2}} \right) = 4cx\)
Contract 6
For each point M in the hyperbola (H), from two equations \(M{F_1}^2 – M{F_2}^2 = 4cx\) and \(\left| {M{F_1} – M{F_2}} \right| = 2a\), prove \(M{F_1} = \left| {a + \frac{c}{a}x} \right| = \left| {a + ex} \right|\) and \(M{F_2} = \left| {a – \frac{c}{a}x} \right| = \left| {a – ex} \right|\)
Detailed explanation:
+ We have: \(M{F_1}^2 – M{F_2}^2 = \left( {M{F_1} – M{F_2}} \right)\left( {M{F_1} + M{F_2} } \right) = \left( {M{F_1} – M{F_2}} \right).\left| {2a} \right| = 4cx\)
\( \Rightarrow M{F_1} – M{F_2} = \frac{{2c}}{{\left| a \right|}}x\)
+ We have: \(\left\{ \begin{array}{l}M{F_1} + M{F_2} = \left| {2a} \right|\quad \left( 1 \right)\\M{ F_1} – M{F_2} = \frac{{2c}}{{\left| a \right|}}x\quad \left( 2 \right)\end{array} \right.\)
From (1) and (2) it follows that:
\(2M{F_1} = \left| {2a} \right| + \frac{{2c}}{{\left| a \right|}}x \Rightarrow M{F_1} = \left| a \right| + \frac{c}{{\left| a \right|}}x = \left| {a + \frac{c}{a}x} \right| = \left| {a + ex} \right| \)
\(M{F_2} = 2\left| a \right| – M{F_1} = 2\left| a \right| – \left( {\left| a \right| + \frac{c}{{\) left| a \right|}}x} \right) = \left| a \right| – \frac{c}{{\left| a \right|}}x\)\( = \left| {a – \frac{c}{a}x} \right| = \left| {a – ex} \right|\)
Practice – use 3
Let the hyperbola (H) have the canonical equation: \(\frac{{{x^2}}}{{144}} – \frac{{{y^2}}}{{25}} = 1\) . Assume that point M is the standard diderm of the hyperbola with a coordinate of 15. Find the lengths of the radii through the focal point of the point M.
Solution method:
Equation of hyperbola \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\” ) where \(a > 0,b > 0\). Then we have:
+ The lengths of the two radii through the focal point of \(M(x;y)\) are: \(M{F_1} = \left| {a + \frac{c}{a}x} \right|;M {F_2} = \left| {a – \frac{c}{a}x} \right|\)
Detailed explanation:
We have \(a = 12,b = 3,c = \sqrt {{a^2} + {b^2}} = \sqrt {144 + 9} = 3\sqrt {17} \).
Thus \(e = \frac{{3\sqrt {17} }}{{12}} = \frac{{\sqrt {17} }}{4}\).
Then the lengths of the radii through the focal point of the point M are:
\(M{F_1} = \left| {12 + \frac{{\sqrt {17} }}{4}.15} \right|;M{F_2} = \left| {12 – \frac{{\ sqrt {17} }}{4}.15} \right|\)
