: Center ( (1, -2) ), ( a^2 = 25 \implies a = 5 ), ( b^2 = 9 \implies b = 3 ). Vertices: ( (1 \pm 5, -2) ) → ( (6, -2) ) and ( (-4, -2) ). ( c = \sqrta^2 - b^2 = \sqrt25 - 9 = 4 ). Foci: ( (1 \pm 4, -2) ) → ( (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 ). ✅ Solved Exercise 10 Find center, vertices, foci of ( \frac(x - 2)^216 - \frac(y + 1)^29 = 1 ).
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
: 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 ). : Center ( (1, -2) ), ( a^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) ). Foci: ( (1 \pm 4, -2) ) → ( (5, -2) ) and ( (-3, -2) )
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 ).