The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 5 } = 1$, is
(1) 27
(2) $\frac { 27 } { 4 }$
(3) 18
(4) $\frac { 27 } { 2 }$

Consider an ellipse, whose center is at the origin and its major axis is along the $x$-axis. If its eccentricity is $\frac { 3 } { 5 }$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is:
(1) 32
(2) 80
(3) 40
(4) 8

Let $S$ and $S ^ { \prime }$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S ^ { \prime } B S$ is a right angled triangle with right angle at $B$ and area $\left( \Delta S ^ { \prime } B S \right) = 8$ sq. units, then the length of a latus rectum of the ellipse is :
(1) $2 \sqrt { 2 }$
(2) 2
(3) 4
(4) $4 \sqrt { 2 }$

The tangent and the normal lines at the point $( \sqrt { 3 } , 1 )$ to the circle $x ^ { 2 } + y ^ { 2 } = 4$ and the $x$-axis form a triangle. The area of this triangle (in square units) is:
(1) $\frac { 1 } { 3 }$
(2) $\frac { 2 } { \sqrt { 3 } }$
(3) $\frac { 4 } { \sqrt { 3 } }$
(4) $\frac { 1 } { \sqrt { 3 } }$

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $( 0 , 5 \sqrt { 3 } )$, then the length of its latus rectum is:
(1) 6
(2) 10
(3) 8
(4) 5

A rectangle is inscribed in a circle with a diameter lying along the line $3 y = x + 7$. If the two adjacent vertices of the rectangle are $( - 8,5 )$ and $( 6,5 )$, then the area of the rectangle (in sq. units) is:
(1) 72
(2) 98
(3) 56
(4) 84

If the curve $x ^ { 2 } + 2 y ^ { 2 } = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$, then the angle subtended by the line segment $PQ$ at the origin is
(1) $\frac { \pi } { 2 } - \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$
(2) $\frac { \pi } { 2 } + \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$
(3) $\frac { \pi } { 2 } + \tan ^ { - 1 } \left( \frac { 1 } { 4 } \right)$
(4) $\frac { \pi } { 2 } - \tan ^ { - 1 } \left( \frac { 1 } { 4 } \right)$

Two tangents are drawn from a point $P$ to the circle $x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 4 = 0$, such that the angle between these tangents is $\tan ^ { - 1 } \left( \frac { 12 } { 5 } \right)$, where $\tan ^ { - 1 } \left( \frac { 12 } { 5 } \right) \in ( 0 , \pi )$. If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\triangle P A B$ and $\triangle C A B$ is:
(1) $11 : 4$
(2) $9 : 4$
(3) $3 : 1$
(4) $2 : 1$

If the tangents drawn at the point $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^2 + y^2 - 2x - 4y = 0$ intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to
(1) $\frac{3+\sqrt{5}}{2}$
(2) $\frac{4+2\sqrt{5}}{2}$
(3) $\frac{5+3\sqrt{5}}{2}$
(4) $\frac{7+3\sqrt{5}}{2}$

Let the tangent to the circle $C _ { 1 } : x ^ { 2 } + y ^ { 2 } = 2$ at the point $M ( - 1,1 )$ intersect the circle $C _ { 2 }$ : $( x - 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 5$, at two distinct points $A$ and $B$. If the tangents to $C _ { 2 }$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $A N B$ is equal to
(1) $\frac { 1 } { 2 }$
(2) $\frac { 2 } { 3 }$
(3) $\frac { 1 } { 6 }$
(4) $\frac { 5 } { 3 }$

The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^2 - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^2$) of the triangle $PAB$ is:
(1) 4
(2) 6
(3) 7
(4) 8

Let $C$ be the centre of the circle $x^2 + y^2 - x + 2y = \frac{11}{4}$ and $P$ be a point on the circle. A line passes through the point $C$, makes an angle of $\frac{\pi}{4}$ with the line $CP$ and intersects the circle at the points $Q$ and $R$. Then the area of the triangle $PQR$ (in unit$^2$) is
(1) 2
(2) $2\sqrt{2}$
(3) $8\sin\frac{\pi}{8}$
(4) $8\cos\frac{\pi}{8}$

A point $P$ moves so that the sum of squares of its distances from the points $( 1,2 )$ and $( - 2,1 )$ is 14. Let $f ( x , y ) = 0$ be the locus of $P$, which intersects the $x$-axis at the points $A , B$ and the $y$-axis at the point $C , D$. Then the area of the quadrilateral $ACBD$ is equal to
(1) $\frac { 9 } { 2 }$
(2) $\frac { 3 \sqrt { 17 } } { 2 }$
(3) $\frac { 3 \sqrt { 17 } } { 4 }$
(4) 9

Let the tangents at two points $A$ and $B$ on the circle $x ^ { 2 } + y ^ { 2 } - 4 x + 3 = 0$ meet at origin $O ( 0,0 )$. Then the area of the triangle of $O A B$ is
(1) $\frac { 3 \sqrt { 3 } } { 2 }$
(2) $\frac { 3 \sqrt { 3 } } { 4 }$
(3) $\frac { 3 } { 2 \sqrt { 3 } }$
(4) $\frac { 3 } { 4 \sqrt { 3 } }$

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2x + y = 5$ meet at the point $R\left(\frac{9}{4}, 2\right)$, then the area of the triangle $PQR$ is
(1) $\frac{5}{4}$
(2) $\frac{13}{8}$
(3) $\frac{5}{8}$
(4) $\frac{13}{4}$

Let $P(a_{1}, b_{1})$ and $Q(a_{2}, b_{2})$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then $a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2}$ is equal to $\_\_\_\_$

Let a circle $C_1$ be obtained on rolling the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ upwards 4 units on the tangent $T$ to it at the point $(3,2)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium AMNB is:
(1) $22 + \sqrt{2}$
(2) $41 + \sqrt{2}$
(3) $3 + 2\sqrt{2}$
(4) $21 + \sqrt{2}$

Let the ellipse $E$: $x^2 + 9y^2 = 9$ intersect the positive $x$- and $y$-axes at the points $A$ and $B$ respectively. Let the major axis of $E$ be a diameter of the circle $C$. Let the line passing through $A$ and $B$ meet the circle $C$ at the point $P$. If the area of the triangle with vertices $A$, $P$ and the origin $O$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m - n$ is equal to
(1) 16
(2) 15
(3) 17
(4) 18

The area (in sq. units) of the part of circle $x ^ { 2 } + y ^ { 2 } = 169$ which is below the line $5 x - y = 13$ is $\frac { \pi \alpha } { 2 \beta } - \frac { 65 } { 2 } + \frac { \alpha } { \beta } \sin ^ { - 1 } \left( \frac { 12 } { 13 } \right)$ where $\alpha , \beta$ are coprime numbers. Then $\alpha + \beta$ is equal to

Let S be the focus of the hyperbola $\frac { x ^ { 2 } } { 3 } - \frac { y ^ { 2 } } { 5 } = 1$, on the positive $x$-axis. Let C be the circle with its centre at $A ( \sqrt { 6 } , \sqrt { 5 } )$ and passing through the point $S$. If $O$ is the origin and $S A B$ is a diameter of $C$, then the square of the area of the triangle OSB is equal to $\_\_\_\_$

Let $A , B$ and $C$ be three points on the parabola $y ^ { 2 } = 6 x$ and let the line segment $A B$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left( \frac { A M \cdot B N } { C D } \right) ^ { 2 }$ is equal to $\_\_\_\_$

Let the line $x + y = 1$ meet the circle $x^2 + y^2 = 4$ at the points A and B. If the line perpendicular to $AB$ and passing through the mid point of the chord $AB$ intersects the circle at $C$ and $D$, then the area of the quadrilateral ADBC is equal to:
(1) $\sqrt{14}$
(2) $3\sqrt{7}$
(3) $2\sqrt{14}$
(4) $5\sqrt{7}$

On the coordinate plane, there is an annular region formed by the intersection of the exterior of the circle $x ^ { 2 } + y ^ { 2 } = 3$ and the interior of the circle $x ^ { 2 } + y ^ { 2 } = 4$ . A person wants to use a straight scanning rod of length 1 to scan a certain region $R$ above the $x$-axis of this annular region. He designs the scanning rod with black and white ends moving respectively on the semicircles $C _ { 1 } : x ^ { 2 } + y ^ { 2 } = 3 ( y \geq 0 )$ and $C _ { 2 } : x ^ { 2 } + y ^ { 2 } = 4 ( y \geq 0 )$ . Initially, the black end of the scanning rod is at point $A ( \sqrt { 3 } , 0 )$ and the white end is at point $B$ on $C _ { 2 }$ . Then the black and white ends move counterclockwise along $C _ { 1 }$ and $C _ { 2 }$ respectively until the white end reaches point $B ^ { \prime } ( - 2,0 )$ on $C _ { 2 }$ , at which point scanning stops.
What are the coordinates of point $B$? (Single-choice question, 3 points)
(1) $( 0,2 )$
(2) $( 1 , \sqrt { 3 } )$
(3) $( \sqrt { 2 } , \sqrt { 2 } )$
(4) $( \sqrt { 3 } , 1 )$
(5) $( 2,0 )$

On the coordinate plane, there is a square and a regular hexagon, with the square to the right of the hexagon. Both regular polygons have one side on the $x$-axis, and the center $A$ of the square and the center $B$ of the hexagon are both above the $x$-axis. The two polygons have exactly one intersection point $P$. The side length of the square is 6, and the distance from point $P$ to the $x$-axis is $2\sqrt{3}$. Select the correct options.
(1) The distance from point $A$ to the $x$-axis is greater than the distance from point $B$ to the $x$-axis
(2) The side length of the regular hexagon is 6
(3) $\overrightarrow{BA} = (7, 3 - 2\sqrt{3})$
(4) $\overline{AP} > \sqrt{10}$
(5) The slope of line $AP$ is greater than $-\frac{1}{\sqrt{3}}$

On the coordinate plane, there is a parallelogram $\Gamma$, where two sides lie on lines parallel to $5x - y = 0$, and the other two sides lie on lines perpendicular to $3x - 2y = 0$. Let $Q$ be the intersection point of the two diagonals of $\Gamma$. It is known that $\Gamma$ has a vertex $P$ satisfying $\overrightarrow{PQ} = (10, -1)$. The area of $\Gamma$ is (11--1)(11--2)(11--3).