Given:

The equation is \( \dfrac{x^2}{36} +\dfrac{ y^2}{16} = 1\)

Here, the denominator of \( \dfrac{x^2}{36}\) is greater than the denominator of \( \dfrac{y^2}{16}\).

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

a = 6 and b = 4.

c = \( \sqrt{(a^2 – b^2)}\)

= \( \sqrt{(36-16)}\)

= \( \sqrt{20}\)

=\( 2\sqrt{5}\)

Then,

The coordinates of the foci are (\( 2\sqrt{5}\), 0) and (\( - 2\sqrt{5}\), 0).

The coordinates of the vertices are (6, 0) and (-6, 0)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, e = \( \dfrac{c}{a}\) = \(\dfrac{2\sqrt{5}}{6} = \sqrt{\dfrac{5}{3}}\)

Length of latus rectum = \( \dfrac{2b^2}{a} = \dfrac{(2×16)}{6}\)

=\( \dfrac{16}{3} \)

Answered by Aaryan | 1 year 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.