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…

Question

Consider the ellipse x 2 4 + y 2 3 = 1 . Let H &#945; , 0 , 0 &#60; &#945; &#60; 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 &#981; with the positive x -axis. List- I List- II I If &#981; = &#960; 4 , then the area of the triangle F G H is P 3 - 1 4 8 II If &#981; = &#960; 3 , then the area of the triangle F G H is Q 1 III If &#981; = &#960; 6 , then the area of the triangle F G H is R 3 4 IV If &#981; = &#960; 12 , then the area of the triangle F G H is S 1 2 3 T 3 3 2 The correct option is:

Aicharya · Accepted Answer

Given x 2 4 + y 2 3 = 1 Let &#945; &#8801; 2 cos &#981; Tangent at E 2 cos &#981; , 3 sin &#981; to the ellipse is x cos &#981; 2 + y sin &#981; 3 = 1 This intersect x -axis at G 2 sec &#981; , 0 Drawing the figure as per the information, we get, Area of triangle F G H = 1 2 2 sec &#981; - 2 cos &#981; 2 sin &#981; &#916; = 2 sin 2 &#981; . tan &#981; &#916; = 1 - cos 2 &#981; &#183; tan &#981; I. If &#981; = &#960; 4 , &#916; = 1 &#8594; Q II. If &#981; = &#960; 3 , &#916; = 2 &#183; 3 2 2 &#183; 3 = 3 3 2 &#8594; T III. If &#981; = &#960; 6 , &#916; = 2 &#183; 1 2 2 &#183; 1 3 = 1 2 3 &#8594; S IV. If &#981; = &#960; 12 , &#916; = 1 - 3 2 &#183; 2 - 3 = 2 - 3 2 2 = 3 - 1 4 8 &#8594; 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	$(S)$	$\frac{1}{2 \sqrt{3}}$
		$(T)$	$\frac{3 \sqrt{3}}{2}$

Consider the ellipse x 2 4 + y 2 3 = 1 . Let H α , 0 , 0 < α < 2 , be a point. A straight…

Solution

Consider the ellipse x 2 4 + y 2 3 = 1 . Let H &#945; , 0 , 0 &#60; &#945; &#60; 2 , be a point. A straight…

Solution

Consider the ellipse x 2 4 + y 2 3 = 1 . Let H α , 0 , 0 < α < 2 , be a point. A straight…