The equation is \( \dfrac{x^2}{100}\) + \( \dfrac{y^2}{400}\) = 1

Here, the denominator of \( \dfrac{y^2}{400}\) is greater than the denominator of \( \dfrac{x^2}{100}\).

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with

\( \dfrac{x^2}{b^2}\) + \( \dfrac{y^2}{a^2}\) = 1, we get

b = 10 and a =20.

c = \( \sqrt{ (a^2 – b^2)}\)

= \( \sqrt{ (400– 100)}\)

= \( \sqrt{ 300}\)

= \(10 \sqrt{3}\)

Then,

The coordinates of the foci are (0,\( 10 \sqrt{3}\)) and (0, \(- 10 \sqrt{3}\)).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

Eccentricity, e =\( \dfrac{c}{a}\)

= \( \dfrac{10\sqrt{3}}{20}\) = \( \dfrac{\sqrt{3}}{2}\)

Length of latus rectum =\( \dfrac{2b^2}{a}\)

= \( \dfrac{(2×10^2)}{20}\) = \( \dfrac{(2×100)}{20}\) = 10

Answered by Abhisek | 2 years agoAn equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.