Geometria Analitica Conamat Ejercicios Resueltos Page

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

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 ). ✅ Solved Exercise 9 Find center, vertices, foci of ( \frac(x - 1)^225 + \frac(y + 2)^29 = 1 ).

: ( (3, 9) ) and ( (-1, 1) ) 8. Parabola Vertex, Focus, Directrix Vertical parabola : ( (x - h)^2 = 4p(y - k) ) Vertex ( (h, k) ), focus ( (h, k + p) ), directrix ( y = k - p ). ✅ Solved Exercise 8 Find vertex, focus, directrix of ( y = 2x^2 - 8x + 5 ).

The article includes theory reminders, step-by-step solved problems, and practical tips. Analytic geometry combines algebra and geometry to study geometric figures using coordinates and equations. It is essential for understanding lines, circles, parabolas, ellipses, and hyperbolas. geometria analitica conamat ejercicios resueltos

: ( y = -3x + 11 ) 5. Equation of a Circle (Center and Radius) Standard form : [ (x - h)^2 + (y - k)^2 = r^2 ] Center ( C(h, k) ), radius ( r ). ✅ Solved Exercise 5 Find the equation of the circle with center ( C(3, -2) ) and radius ( r = 4 ).

: [ (x - 3)^2 + (y + 2)^2 = 16 ]

: 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 ] : [ y - 5 = -3(x -

: [ d = \sqrt(7 - 3)^2 + (5 - 2)^2 = \sqrt4^2 + 3^2 = \sqrt16 + 9 = \sqrt25 = 5 ]

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

: 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 ). : ( (3, 9) ) and ( (-1, 1) ) 8

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

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

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 ] ✅ Solved Exercise 1 Find the distance between ( A(3, 2) ) and ( B(7, 5) ).

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

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