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 ).

: ( (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 ). geometria analitica conamat ejercicios resueltos

: 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 ). : Complete the square: [ y = 2(x^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 ). ✅ Solved Exercise 8 Find vertex, focus, directrix

: ( 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 ).

: ( 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) ).

: 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 ]