Ellipse & Parabola: Equation Modeling and Distance Calculation

This assignment provides a detailed mathematical analysis of ellipses and parabolas, focusing on equation modeling and distance calculations. It begins by determining the equation of an ellipse based on given dimensions, followed by verifying a vertical distance condition. The assignment then explores parabola equations, calculating the general equation and width at a specific height. A comparative analysis between an original and adapted parabolic model highlights differences in height, width, and area coverage, concluding with an assessment of how well an entrance fits within each model. Desklib offers this assignment as a resource for students seeking to understand and solve similar mathematical problems.

1)
Given,
As we can see that the pavilion is in the form of a ellipse, we use elliptical formula,
a) Equation of the ellipse is given by,
x2
a2 + y2
b2 =1
Here a= major axis and b=minor axis.
Here a= 5+7+ 4
2
Which is =8
And b= 2.5+2.5+1
2
Which=3
There fore the equation of the ellipse= x2
82 + y2
32 =1
Which is equal to = x2
64 + y2
9 =1
b) Let the vertical distance from A and D be x
Here, the minor axis =6m
Given the bottom distance =1m.
So by symmetry, the top most distance is also 1m.
The vertical distance (x)= 2.5+2.5+1-(1+1)
That is the distance =4m
Which is greater that 3.6m
There for the given condition is true.
That is it is at least 3.6m.

2)
a)
given maximum height=9.5m and width at ground level = 16m
the equation of a parabola is given by,
y2=−4 ax
Here y=9.5,
x=16
plugging into the parabolic equation, we get
9.52 =−4∗a∗16
a= −1.410
There fore the general equation of the parabola is,
y2=−1.410 x
b)
width of the building at a height of 3.6m
this is formed by substituting y=3.6 at the parabolic equation.
That is y=3.6 in y2=¿-1.410x
The solution for this equation is x=9.19m

3)
Given, the vertical height of the parabola = 3.6m at the same width as that of ellipse.
y=3.6m, x is to find out.
the equation of a parabola is given by,
y2=−4 ax
Where a=-1.410
Substituting in the parabolic equation,
3.62 =−4∗−1.410∗x
x=2.297m
plugging into the parabolic equation, we get
y2=−4 ax
3.62 =−4∗−1.410∗2.297
The difference between original model and adapted model is,
That is the original model has larger height and larger width than the adapted model.
That is area covered is more for original than the adapted model.
The entrance perfectly fits inside the original parabola where as it doesn’t fit in the adapted model.