Consider the ellipse x 2 4 + y 2 3 = 1 . Let H α , 0 , 0

Question

Consider the ellipse x 2 4 + y 2 3 = 1 . Let H α , 0 , 0 < α < 2 , be a point. A straight line drawn through H parallel to the y -axis crosses the ellipse and its auxiliary circle at points E and F respectively, in the first quadrant. The tangent to the ellipse at the point E intersects the positive x -axis at a point G . Suppose the straight line joining F and the origin makes an angle ϕ with the positive x -axis. List- I List- II I If ϕ = π 4 , then the area of the triangle F G H is P 3 - 1 4 8 II If ϕ = π 3 , then the area of the triangle F G H is Q 1 III If ϕ = π 6 , then the area of the triangle F G H is R 3 4 IV If ϕ = π 12 , then the area of the triangle F G H is S 1 2 3 T 3 3 2 The correct option is:

TestHub · Accepted Answer

Given x 2 4 + y 2 3 = 1 Let α ≡ 2 cos ϕ Tangent at E 2 cos ϕ , 3 sin ϕ to the ellipse is x cos ϕ 2 + y sin ϕ 3 = 1 This intersect x -axis at G 2 sec ϕ , 0 Drawing the figure as per the information, we get, Area of triangle F G H = 1 2 2 sec ϕ - 2 cos ϕ 2 sin ϕ Δ = 2 sin 2 ϕ . tan ϕ Δ = 1 - cos 2 ϕ · tan ϕ I. If ϕ = π 4 , Δ = 1 → Q II. If ϕ = π 3 , Δ = 2 · 3 2 2 · 3 = 3 3 2 → T III. If ϕ = π 6 , Δ = 2 · 1 2 2 · 1 3 = 1 2 3 → S IV. If ϕ = π 12 , Δ = 1 - 3 2 · 2 - 3 = 2 - 3 2 2 = 3 - 1 4 8 → P

	List- $I$		List- $II$
$(I)$	If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	$(P)$	$\frac{{(\sqrt{3} - 1)}^{4}}{8}$
$(II)$	If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is	$(Q)$	$1$
$(III)$	If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is	$(R)$	$\frac{3}{4}$
$(IV)$	If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is	$(S)$	$\frac{1}{2 \sqrt{3}}$
		$(T)$	$\frac{3 \sqrt{3}}{2}$

Mathematics - Ellipse Question with Solution | TestHub

Options:

Doubts & Discussion