## Solution Section 4 Page 53 Math Learning Topic 10 – Kite>

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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:

$$M{F_1} = \left| {12 + \frac{{\sqrt {17} }}{4}.15} \right|;M{F_2} = \left| {12 – \frac{{\ sqrt {17} }}{4}.15} \right|$$