As we know that: For a given ellipse we can give the coordinates of a point on the given ellipse by (acos θ, bsin θ)
From figure we can say that length of rectangle = 2a cosθ
Note: In reference to figure - (taking z = θ)
And breadth of inscribed rectangle = 2b sinθ
Let A be the area of rectangle inscribed.
∴ A = 4ab sinθ cos θ = 2ab sin 2θ
We need to maximise the area.
⇒ 4ab cos 2θ = 0
⇒ cos 2θ = 0
∴ θ = π/4
∴ θ = π/4 is the point of maxima.
Hence area of the greatest rectangle that can be inscribed in the given ellipse is given by
A = 2ab sin(2×π/4) = 2ab.
Given equation is 3x2 – y2 = 8.
To find the equation of tangent we need to find the slope first and the slope is given by the value of derivative at that point.
As, 3x2 – y2 = 8
Differentiating both sides w.r.t x, we get –
As slope comes to be infinite ⇒ line is parallel to y – axis
Given that it passes through (4/3,0)
∴ line parallel to y-axis and passing through (4/3,0) is
x = 4/3
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