Geometria - Analitica Conamat Ejercicios Resueltos

: [ M_x = \frac-2 + 62 = \frac42 = 2, \quad M_y = \frac4 + (-8)2 = \frac-42 = -2 ]

Below, you will find covering the most common topics, explained step by step. 1. Distance Between Two Points Formula : [ d = \sqrt(x_2 - x_1)^2 + (y_2 - y_1)^2 ] Б°┘ Solved Exercise 1 Find the distance between ( A(3, 2) ) and ( B(7, 5) ).

Vertex ( (2, -3) ), focus ( (2, -3 + 1/8) = (2, -23/8) ), directrix ( y = -3 - 1/8 = -25/8 ). Equation : [ \frac(x - h)^2a^2 + \frac(y - k)^2b^2 = 1, \quad a > b ] Center ( (h, k) ), vertices ( (h \pm a, k) ), foci ( (h \pm c, k) ), ( c^2 = a^2 - b^2 ). Б°┘ Solved Exercise 9 Find center, vertices, foci of ( \frac(x - 1)^225 + \frac(y + 2)^29 = 1 ).

: [ m = \frac9 - 34 - 1 = \frac63 = 2 ]

: [ y - 5 = -3(x - 2) \implies y - 5 = -3x + 6 \implies y = -3x + 11 ]

: Group ( x ) and ( y ) terms: [ (x^2 - 6x) + (y^2 + 4y) = 3 ] Complete the square: [ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4 ] [ (x - 3)^2 + (y + 2)^2 = 16 ] Center ( C(3, -2) ), radius ( r = 4 ). 7. Intersection of a Line and a Parabola Б°┘ Solved Exercise 7 Find intersection points between ( y = x^2 ) and ( y = 2x + 3 ).

: Center ( (1, -2) ), ( a^2 = 25 \implies a = 5 ), ( b^2 = 9 \implies b = 3 ). Vertices: ( (1 \pm 5, -2) ) Б├▓ ( (6, -2) ) and ( (-4, -2) ). ( c = \sqrta^2 - b^2 = \sqrt25 - 9 = 4 ). Foci: ( (1 \pm 4, -2) ) Б├▓ ( (5, -2) ) and ( (-3, -2) ). 10. Hyperbola (Horizontal Transverse Axis) Equation : [ \frac(x - h)^2a^2 - \frac(y - k)^2b^2 = 1 ] Center ( (h, k) ), vertices ( (h \pm a, k) ), foci ( (h \pm c, k) ), ( c^2 = a^2 + b^2 ). Б°┘ Solved Exercise 10 Find center, vertices, foci of ( \frac(x - 2)^216 - \frac(y + 1)^29 = 1 ). geometria analitica conamat ejercicios resueltos

: ( (x - 3)^2 + (y + 2)^2 = 16 ) 6. Circle from General Form to Standard Form Б°┘ Solved Exercise 6 Convert ( x^2 + y^2 - 6x + 4y - 3 = 0 ) to standard form and find center and radius.

: Complete the square: [ y = 2(x^2 - 4x) + 5 = 2(x^2 - 4x + 4 - 4) + 5 ] [ y = 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3 ] Rewrite: [ y + 3 = 2(x - 2)^2 \implies (x - 2)^2 = \frac12(y + 3) ] So ( 4p = \frac12 \implies p = \frac18 ).

: ( M(2, -2) ) 3. Slope of a Line Formula : [ m = \fracy_2 - y_1x_2 - x_1 ] Б°┘ Solved Exercise 3 Find the slope through ( A(1, 3) ) and ( B(4, 9) ). : [ M_x = \frac-2 + 62 =

: ( d = 5 ) 2. Midpoint of a Segment Formula : [ M = \left( \fracx_1 + x_22, \fracy_1 + y_22 \right) ] Б°┘ Solved Exercise 2 Find the midpoint of ( P(-2, 4) ) and ( Q(6, -8) ).

: Set equal: [ x^2 = 2x + 3 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0 ] [ x = 3 \implies y = 9 \quad \textand \quad x = -1 \implies y = 1 ]

: ( m = 2 ) 4. Equation of a Line (Point-Slope Form) Formula : [ y - y_1 = m(x - x_1) ] Б°┘ Solved Exercise 4 Find the line equation with slope ( m = -3 ) passing through ( (2, 5) ). Vertex ( (2, -3) ), focus ( (2,