Let $F _ { 1 } \left( x _ { 1 } , 0 \right)$ and $F _ { 2 } \left( x _ { 2 } , 0 \right)$, for $x _ { 1 } < 0$ and $x _ { 2 } > 0$, be the foci of the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 8 } = 1$. Suppose a parabola having vertex at the origin and focus at $F _ { 2 }$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
The orthocentre of the triangle $F _ { 1 } M N$ is
(A) $\left( - \frac { 9 } { 10 } , 0 \right)$
(B) $\left( \frac { 2 } { 3 } , 0 \right)$
(C) $\left( \frac { 9 } { 10 } , 0 \right)$
(D) $\left( \frac { 2 } { 3 } , \sqrt { 6 } \right)$

Let $F _ { 1 } \left( x _ { 1 } , 0 \right)$ and $F _ { 2 } \left( x _ { 2 } , 0 \right)$, for $x _ { 1 } < 0$ and $x _ { 2 } > 0$, be the foci of the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 8 } = 1$. Suppose a parabola having vertex at the origin and focus at $F _ { 2 }$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F _ { 1 } N F _ { 2 }$ is
(A) $3 : 4$
(B) $4 : 5$
(C) $5 : 8$
(D) $2 : 3$

If $2x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$, then which of the following CANNOT be sides of a right angled triangle?
[A] $a, 4, 1$
[B] $a, 4, 2$
[C] $2a, 8, 1$
[D] $2a, 4, 1$

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

	Column 1	Column 2	Column 3
	(I) $x^2 + y^2 = a^2$	(i) $my = m^2x + a$	(P) $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$
(II)	$x^2 + a^2y^2 = a^2$	(ii) $y = mx + a\sqrt{m^2+1}$	(Q) $\left(\frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right)$
(III)	$y^2 = 4ax$	(iii) $y = mx + \sqrt{a^2m^2 - 1}$	(R) $\left(\frac{-a^2m}{\sqrt{a^2m^2+1}}, \frac{1}{\sqrt{a^2m^2+1}}\right)$
(IV)	$x^2 - a^2y^2 = a^2$	(iv) $y = mx + \sqrt{a^2m^2+1}$	(S) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$

The tangent to a suitable conic (Column 1) at $\left(\sqrt{3}, \frac{1}{2}\right)$ is found to be $\sqrt{3}x + 2y = 4$, then which of the following options is the only CORRECT combination?
[A] (IV) (iii) (S)
[B] (IV) (iv) (S)
[C] (II) (iii) (R)
[D] (II) (iv) (R)

Let $H$ : $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, where $a > b > 0$, be a hyperbola in the $x y$-plane whose conjugate axis $L M$ subtends an angle of $60 ^ { \circ }$ at one of its vertices $N$. Let the area of the triangle $L M N$ be $4 \sqrt { 3 }$.
LIST-I P. The length of the conjugate axis of $H$ is Q. The eccentricity of $H$ is R. The distance between the foci of $H$ is S. The length of the latus rectum of $H$ is
LIST-II

1. 8
2. $\frac { 4 } { \sqrt { 3 } }$
3. $\frac { 2 } { \sqrt { 3 } }$
4. 4

The correct option is:
(A) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 2 } ; \mathbf { R } \rightarrow \mathbf { 1 } ; \mathbf { S } \rightarrow \mathbf { 3 }$
(B) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 3 } ; \mathbf { R } \rightarrow \mathbf { 1 } ; \mathbf { S } \rightarrow \mathbf { 2 }$
(C) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 1 } ; \mathbf { R } \rightarrow \mathbf { 3 } ; \mathbf { S } \rightarrow \mathbf { 2 }$
(D) $\mathbf { P } \rightarrow \mathbf { 3 } ; \mathbf { Q } \rightarrow \mathbf { 4 } ; \mathbf { R } \rightarrow \mathbf { 2 } ; \mathbf { S } \rightarrow \mathbf { 1 }$

Let $a , b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y ^ { 2 } = 4 \lambda x$, and suppose the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ passes through the point $P$. If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other, then the eccentricity of the ellipse is
(A) $\frac { 1 } { \sqrt { 2 } }$
(B) $\frac { 1 } { 2 }$
(C) $\frac { 1 } { 3 }$
(D) $\frac { 2 } { 5 }$

Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$. Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. Suppose the tangent to the hyperbola at $P$ passes through the point $(1, 0)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$, the normal at $P$ and the $x$-axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?
(A) $1 < e < \sqrt{2}$
(B) $\sqrt{2} < e < 2$
(C) $\Delta = a^{4}$
(D) $\Delta = b^{4}$

Consider the hyperbola
$$\frac { x ^ { 2 } } { 100 } - \frac { y ^ { 2 } } { 64 } = 1$$
with foci at $S$ and $S _ { 1 }$, where $S$ lies on the positive $x$-axis. Let $P$ be a point on the hyperbola, in the first quadrant. Let $\angle S P S _ { 1 } = \alpha$, with $\alpha < \frac { \pi } { 2 }$. The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S _ { 1 } P$ at $P _ { 1 }$. Let $\delta$ be the distance of $P$ from the straight line $S P _ { 1 }$, and $\beta = S _ { 1 } P$. Then the greatest integer less than or equal to $\frac { \beta \delta } { 9 } \sin \frac { \alpha } { 2 }$ is $\_\_\_\_$ .

Let $P$ be a point on the parabola $y ^ { 2 } = 4 a x$, where $a > 0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $P F Q$, where $F$ is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $( a , m )$ is
(A) $( 2,3 )$
(B) $( 1,3 )$
(C) $( 2,4 )$
(D) $( 3,4 )$

Consider the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1$. Let $S ( p , q )$ be a point in the first quadrant such that $\frac { p ^ { 2 } } { 9 } + \frac { q ^ { 2 } } { 4 } > 1$. Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac { 3 } { 2 }$, then which of the following options is correct?
(A) $q = 2 , p = 3 \sqrt { 3 }$
(B) $q = 2 , p = 4 \sqrt { 3 }$
(C) $q = 1 , p = 5 \sqrt { 3 }$
(D) $q = 1 , p = 6 \sqrt { 3 }$

For the hyperbola $\frac { x ^ { 2 } } { \cos ^ { 2 } \alpha } - \frac { y ^ { 2 } } { \sin ^ { 2 } \alpha } = 1$, which of the following remains constant when $\alpha$ varies?
(1) eccentricity
(2) directrix
(3) abscissae of vertices
(4) abscissae of foci

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$ is
(1) $5x^{2}+3y^{2}-48=0$
(2) $3x^{2}+5y^{2}-15=0$
(3) $5x^{2}+3y^{2}-32=0$
(4) $3x^{2}+5y^{2}-32=0$

The area of triangle formed by the lines joining the vertex of the parabola, $x^{2} = 8y$, to the extremities of its latus rectum is
(1) 2
(2) 8
(3) 1
(4) 4

An ellipse is drawn by taking a diameter of the circle $(x-1)^{2}+y^{2}=1$ as its semi-minor axis and a diameter of the circle $x^{2}+(y-2)^{2}=4$ as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is
(1) $4x^{2}+y^{2}=4$
(2) $x^{2}+4y^{2}=8$
(3) $4x^{2}+y^{2}=8$
(4) $x^{2}+4y^{2}=16$

If $P_{1}$ and $P_{2}$ are two points on the ellipse $\frac{x^{2}}{4} + y^{2} = 1$ at which the tangents are parallel to the chord joining the points $(0, 1)$ and $(2, 0)$, then the distance between $P_{1}$ and $P_{2}$ is
(1) $2\sqrt{2}$
(2) $\sqrt{5}$
(3) $2\sqrt{3}$
(4) $\sqrt{10}$

Statement 1: $y = m x - \frac { 1 } { m }$ is always a tangent to the parabola, $y ^ { 2 } = - 4 x$ for all non-zero values of $m$. Statement 2: Every tangent to the parabola, $y ^ { 2 } = - 4 x$ will meet its axis at a point whose abscissa is nonnegative.
(1) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1.
(2) Statement 1 is false, Statement 2 is true.
(3) Statement 1 is true, Statement 2 is false.
(4) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

The equation of the normal to the parabola, $x ^ { 2 } = 8 y$ at $x = 4$ is
(1) $x + 2 y = 0$
(2) $x + y = 2$
(3) $x - 2 y = 0$
(4) $x + y = 6$

If the eccentricity of a hyperbola $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, which passes through $( K , 2 )$, is $\frac { \sqrt { 13 } } { 3 }$, then the value of $K ^ { 2 }$ is
(1) 18
(2) 8
(3) 1
(4) 2

The chord $PQ$ of the parabola $y ^ { 2 } = x$, where one end $P$ of the chord is at point $( 4 , - 2 )$, is perpendicular to the axis of the parabola. Then the slope of the normal at $Q$ is
(1) $-4$
(2) $- \frac { 1 } { 4 }$
(3) $4$
(4) $\frac { 1 } { 4 }$

The normal at $\left( 2 , \frac { 3 } { 2 } \right)$ to the ellipse, $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 3 } = 1$ touches a parabola, whose equation is
(1) $y ^ { 2 } = - 104 x$
(2) $y ^ { 2 } = 14 x$
(3) $y ^ { 2 } = 26 x$
(4) $y ^ { 2 } = - 14 x$

The point of intersection of the normals to the parabola $y ^ { 2 } = 4 x$ at the ends of its latus rectum is :
(1) $( 0,2 )$
(2) $( 3,0 )$
(3) $( 0,3 )$
(4) $( 2,0 )$

A tangent to the hyperbola $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 2 } = 1$ meets $x$-axis at P and $y$-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on :
(1) $\frac { 4 } { x ^ { 2 } } + \frac { 2 } { y ^ { 2 } } = 1$
(2) $\frac { 2 } { x ^ { 2 } } - \frac { 4 } { y ^ { 2 } } = 1$
(3) $\frac { 2 } { x ^ { 2 } } + \frac { 4 } { y ^ { 2 } } = 1$
(4) $\frac { 4 } { x ^ { 2 } } - \frac { 2 } { y ^ { 2 } } = 1$

Equation of the line passing through the points of intersection of the parabola $x ^ { 2 } = 8 y$ and the ellipse $\frac { x ^ { 2 } } { 3 } + y ^ { 2 } = 1$ is :
(1) $y - 3 = 0$
(2) $y + 3 = 0$
(3) $3 y + 1 = 0$
(4) $3 y - 1 = 0$

A chord is drawn through the focus of the parabola $y ^ { 2 } = 6 x$ such that its distance from the vertex of this parabola is $\frac { \sqrt { 5 } } { 2 }$, then its slope can be
(1) $\frac { \sqrt { 5 } } { 2 }$
(2) $\frac { 2 } { \sqrt { 3 } }$
(3) $\frac { \sqrt { 3 } } { 2 }$
(4) $\frac { 2 } { \sqrt { 5 } }$

The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 5 } = 1$, meets the $x$-axis and $y$-axis at $A$ and $B$, respectively. Then $OA ^ { 2 } - OB ^ { 2 }$, where $O$ is the origin, equals
(1) $- \frac { 20 } { 9 }$
(2) $\frac { 16 } { 9 }$
(3) 4
(4) $- \frac { 4 } { 3 }$