Title: Sixth Partial Exam
Course: 2025 Probabilidad y Estadística

Q: Given the ellipse  \( \frac{x^2}{49}+ \frac{(y-5)^2}{81}=1 \), the alignment is [...], the center is [...], a = [...], b=[...], c=[...]. The [...] is \( \frac{ 4\sqrt{2} }{9} \) and the [...] is \( \frac{98}{9} \).
Also offered: horizontal; (0,-5); major axis; focal axis; (5,0); minor axis; √130; foci
A: vertical; (0,5); 9; 7; √32; eccentricity; Latus Rectum

Q: Determine the equation of the ellipse with center at the origin and V(0,7) and F(0,5).
- \( 49x^2+24y^2-1=0 \)
- \( 24x^2+49y^2-1=0 \)
- \( 49x^2+24y^2-1176=0 \)
- \( 24x^2+49y^2-1176=0 \)
A: \( 49x^2+24y^2-1176=0 \)

Q: Determine the equation of the horizontal ellipse with center at (1,-1) if the length of the major axis is 10 and \( e= \frac{3}{5} \).
- \(16x^2+25y^2-32x+50y-359=0\)
- \(16x^2+25y^2-32x+50y-40=0\)
- \(16x^2+25y^2+32x-50y-359=0\)
- \(16x^2+25y^2+32x-50y-40=0\)
A: \(16x^2+25y^2-32x+50y-359=0\)

Q: Given the ellipse \( x^2+100y^2-6x-1000y+2409=0 \), determine the value of a. Type your answer in decimal, rounded to the nearest hundreth, if needed. 
A: 10

Q: Given the ellipse \( x^2+100y^2-6x-1000y+2409=0 \), determine the value of b.  Type your answer in decimal, rounded to the nearest hundreth, if needed. 
A: 1

Q: Given the ellipse \( x^2+100y^2-6x-1000y+2409=0 \), determine the length of Latus Rectum. Type your answer in decimal, rounded to the nearest hundreth, if needed. 
A: 0.2

Q: Given the ellipse \( 64x^2+28y^2-1792=0 \), determine the eccentricity. Type your answer in decimal, rounded to the nearest hundreth, if needed.
A: 0.75

Q: Given the ellipse \( 64x^2+28y^2-1792=0 \), determine the value of b.  Type your answer in decimal, rounded to the nearest hundreth, if needed.
A: 5.29

Q: Determine the center of the ellipse whose equation is \( 4x^2+5y^2-16x+30y-4=0 \).  Input your answer in the form (h,k) or no credti will be given.
A: (2,-3) / (2, -3)