Geometria Analitica Conamat Ejercicios Resueltos Now

: ( 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) ).

: 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 ). geometria analitica conamat ejercicios resueltos

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. : ( m = 2 ) 4

: 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 ). Vertices: ( (1 \pm 5, -2) ) → ( (6, -2) ) and ( (-4, -2) )

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

: ( 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 ).