Geometria Analitica Conamat Ejercicios Resueltos Apr 2026
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 ).
: ( M(2, -2) ) 3. Slope of a Line Formula : [ m = \fracy_2 - y_1x_2 - x_1 ] ✅ Solved Exercise 3 Find the slope through ( A(1, 3) ) and ( B(4, 9) ). geometria analitica conamat ejercicios resueltos
: ( 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) ). Vertex ( (2, -3) ), focus ( (2,
: [ y - 5 = -3(x - 2) \implies y - 5 = -3x + 6 \implies y = -3x + 11 ] : ( M(2, -2) ) 3
: ( 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 ).
: 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 ).
: 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 ✅ Solved Exercise 7 Find intersection points between ( y = x^2 ) and ( y = 2x + 3 ).




