In an ellipse 4x²+9y²=144 inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis. Longer side which is parallel to major axis, relates to the shorter sides as 3:2. Find area of rectangle.

This doesn't make sense! If the ellipse is inscribed in the in the rectangle then the vertices of the rectangle would be outside the ellipse. I assume you mean that the rectangle is inscribed in the ellipse (or, equivalently, that the ellipse circumscribes the rectangle).

If that is the case, and the ratio of longer side of the rectangle to shorter side is 3 to 2, we can take the shorter side to be "2a" and the length of the longer side to be "3a". The area of the rectangle is [tex](2a)(3a)= 6a^2[/tex]. We need to find "[tex]a^2[/tex]".

The ellipse is [tex]4x^2+ 9y^2= 144[/tex]. When y= 0, [tex]4x^2= 144[/tex] so [tex]x^2= 36[/tex], [tex]x= \pm 6[/tex] so two vertices are (-6, 0) and (6, 0). When x= 0, [tex]9y^2= 144[/tex] so [tex]y^2= 16[/tex]. [tex]y= \pm 4[/tex] so the other two vertices are (0, -4) and (0, 4). Since the major axis is on the x-axis, we have the longer side of the inscribed rectangle parallel to the x-axis and the shorter side parallel to the y-axis. The vertex of the rectangle in the first quadrant will be [tex](3a/2, a)[/tex] and must satisfy [tex]4(3a/2)^2+9a^2= 9a^2+ 9a^2= 18a^2= 144[/tex]. [tex]a^2= \frac{144}{18}= 8[/tex].

The area of the rectangle is 6(8)= 48.