The word problems in conic sections meant for the application problems in the analytical geometry based on the conic sections like ellipse, parabola, hyperbola, etc..The word problems generally does not give any equations and any portion that does not explain any part which conic sections are present in the concerned problem i.e ellipse, parabola or hyperbola. We have to think it own for the problem exist in the words.
(source from Wikipedia)
word problems conic sections
For the conic section parabola
- consider we have the telescope which is in the form of reflecting have a mirror in parabolic fom for which the vertex to the focus distance is 9mts. If the distance across (in diameter) the top of the mirror is 160cm, find how depth is the mirror at the middle part.
Considering the vertex at the origin.
The VF distance is given by a = 900
The parabola’s equation is y2 = 4 × 900 × x
considering the point ‘x1′ as the mirror’s depth at the middle part
We know that the point (x1, 80) lies on the parabola
802 = 4 × 900 × x1 ⇒ x1 =16/9
∴The mirror’s depth =16/9 cm
For the conic section ellipse
- consider an arch which is in the form of a ellipse halfly(i.e in the form of semi-ellipse) whose span is about 48 feet wide. The arch’s height is about 20 feet. what is the arch’s wide at a height of 10 feet above the base
As the initial step we have to take the mid point of the base as the centre C (0, 0)
Here the base width is about 48 feet, the vertices A and A′ are (24, 0) and (− 24, 0) respectively.
Here we know that 2a = 48 and b = 20.
The corresponding equation is`x^2/24^2` + `y^2/20^2` = 1
consider x1 is the distance from the pole whose height is 10m to the centre.
Then (x1, 10) satisfies the equation (1)
`(x1)^2/24^2` + `10^2/20^2` = 1⇒ x1 = 12`sqrt(3)`
Here the arch’s wide is at a height of 10 feet is 2×1 = 24`sqrt(3)`
Thus the required width of arch is 24`sqrt(3)` feet.
More word problems in conic sections.
For the conic section hyperbola
Find the hyperbola’s equation whose focus on both the sides are (0, ± 5) and the transverse axes length is 6.
From the given data the transverse axis is along y-axis and hence the equation is of the form
`(y-k)^2/a^2` – `(x-h)^2/b^2` = 1
For the centre, C (h, k) is the midpoint of F1 and F2
i.e. C is(`(0+0)/2` , `(5-5)/2` ) = (0,0)
F1F2 = 2ae = 10
The transverse axes length is given by `=>` 2a = 6
⇒ a = 3 and e = 5/3
b2 = a2 (e2 − 1)
∴ The hyperbola’s equation is `y^2/9` – `x^2/16` = 1