(^O^|||看到即做)輕鬆StupidMaths=[好趕]

This is a Pure Maths topic about 2D Co-Geo.


(1) xy = c^2

When you refer to the formula, we have c^2 = 4

c = 2 or c = -2

Hence, the two vertices are ( 2, 2 ) and ( -0, -2 )

The two asymptotes are definitely two axes by definition, which are x = 0 and y = 0 respectively.

We let the foci be ( m, m )

By Pythagoras Theorem, we have


m^2 = ( +/- 2 )^2 + ( +/- 2 )^2 = 8

m = 2 Sq. Root ( 2 ) or m = - 2 Sq. Root ( 2 )

Hence, two foci are ( 2 Sq. Root ( 2 ), 2 Sq. Root ( 2 ) ) and ( - 2 Sq. Root ( 2 ), - 2 Sq. Root ( 2 ) )


(2) Sounds it is a normal hyperbola, not the rectangular hyperbola for the product of x-y coordinates are negative.

General equation: x^2 / a^2 - y^2 / b^2 = 1

Put ( x, y ) = ( 4, -3 ), we have

16 / a^2 - 9 / b^2 = 1

Since two axes are both axes of symmetry, we have a^2 = b^2.

Hence, 16 / a^2 - 9 / a^2 = 1

7 / a^2 = 1

a^2 = 7

Hence, b^2 = 7

The equation of hyperbola is thus

x^2 / 7 - y^2 / 7 = 1

x^2 - y^2 = 7


(3) Put ( x, y ) = ( 5, 4 ) we have

k^2 = ( 5 )( 4 ) = 20

k = 2 Sq. Root ( 5 ) or k = - 2 Sq. Root ( 5 )

Hence, the 2 vertices are ( 2 Sq. Root ( 5 ), 2 Sq. Root ( 5 ) ) and ( - 2 Sq. Root ( 5 ), - 2 Sq. Root ( 5 ) )

We let the foci be ( m, m )

By Pythagoras Theorem, we have

m^2 = ( +/- 2 Sq. Root ( 5 ) )^2 + ( +/- 2 Sq. Root ( 5 ) )^2 = 40

m = 2 Sq. Root ( 10 ) or m = - 2 Sq. Root ( 10 )

Hence, two foci are ( 2 Sq. Root ( 10 ), 2 Sq. Root ( 10 ) ) and ( - 2 Sq. Root ( 2 ), - 2 Sq. Root ( 2 ) )

Hence, two foci are ( 2 Sq. Root ( 2 ), 2 Sq. Root ( 2 ) ) and ( - 2 Sq. Root ( 10 ), - 2 Sq. Root ( 10 ) )


Hope I can help you.