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**Topic**

In the coordinate plane Oxy, for the hyperbola the canonical equation is \(\frac{{{x^2}}}{4} – \frac{{{y^2}}}{1} = 1\)

a) Determine the coordinates of the vertices, the focal point, the focal length, the real axis length of the hyperbola

b) Determine the equations of the asymptotes of the hyperbola and draw the hyperbola above.

**Solution method – See details**

Equation of hyperbola \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\” ) where \(a > 0,b > 0\). Then we have:

+ Focus \({F_1}( – c;0),{F_2}(c;0)\)

+ The vertices are \({A_1}\left( { – a;0} \right),{A_2}\left( {a;0} \right)\)

+ Focal length: \(2c = 2\sqrt {{a^2} + {b^2}} \)

+ Real axis length: \(2a\), imaginary axis length: \(2b\)

+ The two asymptotes of the hyperbola (H) have the equation \(y = – \frac{b}{a}x,y = \frac{b}{a}x\)

**Detailed explanation**

a) We have: \(a = 2,b = 1 \Rightarrow c = \sqrt {{a^2} + {b^2}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \)

+ The coordinates of the hyperbola’s vertices are \({A_1}\left( { – 2;0} \right),{A_2}\left( {2;0} \right)\)

+ The coordinates of the hyperbola’s focal points are \({F_1}( – \sqrt 5 ;0),{F_2}(\sqrt 5 ;0)\)

+ The focal length of the hyperbola is \(2c = 2\sqrt 5 \)

+ Real axis length: \(2a = 4\), imaginary axis length: \(2b = 2\)

b) We have the equations of the asymptotes of the hyperbola (H) with the equation \(y = – \frac{1}{2}x,y = \frac{1}{2}x\)

Draw the hyperbola (H):

We see \(a = 2,b = 1\). (H) has vertices \({A_1}\left( { – 2;0} \right),{A_2}\left( {2;0} \right)\)

Step 1: Draw a basic rectangle with four sides belonging to four straight lines \(x = – 2,x = 2,y = – 1,y = 1\)

Step 2: Draw two diagonals of the base rectangle

Find some specific points on the hyperbola, for example, we find that the point \(M\left( {3;\frac{9}{4}} \right)\) belongs to (H) and the point \({M_1}\ left( {3; – \frac{9}{4}} \right),{M_2}\left( { – 3;\frac{9}{4}} \right),{M_3}\left( { – 3; – \frac{9}{4}} \right)\) belongs to (H)

Step 3: Draw a hyperbolic line (H) outside the base rectangle, the left branch touches the base rectangle’s edge at the point \({A_1}\left( { – 2;0} \right)\ ) and the point \({M_2},{M_3}\); the right branch tangent to the base rectangle’s edge at points \({A_2}\left( {2;0} \right)\) and points \(M,{M_1}\). Draw points on the hyperbola as far from the origin as possible to the asymptote. Hyperbola takes the origin as the center of symmetry and the two coordinate axes as two axes of symmetry.

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