The equation is 16x2 – 9y2 = 576
Let us divide the whole equation by 576, we get
\( \dfrac{16x^2}{576}- \dfrac{9y^2}{576}\)
= \( \dfrac{576}{576}\)
\( \dfrac{x^2}{36}- \dfrac{y^2}{64}=1\)
On comparing this equation with the standard equation of hyperbola
\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1,
We get a = 6 and b = 8,
It is known that, a2 + b2 = c2
So,
c2 = 36 + 64
c2 = \( \sqrt{100}\)
c = 10
Then,
The coordinates of the foci are (10, 0) and (-10, 0).
The coordinates of the vertices are (6, 0) and (-6, 0).
Eccentricity, e = \( \dfrac{c}{a}\)
= \( \dfrac{10}{6}\) =\( \dfrac{5}{3}\)
Length of latus rectum = \( \dfrac{2b^2}{a}\)
= \( \dfrac{(2×8^2)}{6}\)
= \( \dfrac{(2×64)}{6}\)
= \( \dfrac{64}{3}\)
Answered by Abhisek | 1 year agoAn equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
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