The eccentricity of an ellipse whose centre is at the origin is $\dfrac{1}{2}$. If one of its directrices is $x = -4$, then the equation of the normal to it at $\left(1, \dfrac{3}{2}\right)$ is:
(1) $2y - x = 2$
(2) $4x - 2y = 1$
(3) $4x + 2y = 7$
(4) $x + 2y = 4$

A hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the tangent to this hyperbola at $P$ also passes through the point
(1) $(3\sqrt{2}, 2\sqrt{3})$
(2) $(2\sqrt{2}, 3\sqrt{3})$
(3) $(\sqrt{3}, \sqrt{2})$
(4) $(-\sqrt{2}, -\sqrt{3})$

The equation of a tangent to the hyperbola, $4 x ^ { 2 } - 5 y ^ { 2 } = 20$, parallel to the line $x - y = 2$, is
(1) $x - y + 7 = 0$
(2) $x - y - 3 = 0$
(3) $x - y + 1 = 0$
(4) $x - y + 9 = 0$

If the line $y = m x + 7 \sqrt { 3 }$ is normal to the hyperbola $\frac { x ^ { 2 } } { 24 } - \frac { y ^ { 2 } } { 18 } = 1$, then a value of $m$ is:
(1) $\frac { \sqrt { 5 } } { 2 }$
(2) $\frac { 3 } { \sqrt { 5 } }$
(3) $\frac { \sqrt { 15 } } { 2 }$
(4) $\frac { 2 } { \sqrt { 5 } }$

If $y = m x + 4$ is a tangent to both the parabolas, $y ^ { 2 } = 4 x$ and $x ^ { 2 } = 2 b y$, then $b$ is equal to
(1) $-32$
(2) $-64$
(3) $-128$
(4) 128

If a hyperbola passes through the point $P(10, 16)$, and it has vertices at $(\pm 6, 0)$, then the equation of the normal to it at $P$ is.
(1) $3x + 4y = 94$
(2) $2x + 5y = 100$
(3) $x + 2y = 42$
(4) $x + 3y = 58$

A line parallel to the straight line $2x - y = 0$ is tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$. Then $x_{1}^{2} + 5y_{1}^{2}$ is equal to
(1) 6
(2) 8
(3) 10
(4) 5

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac { 1 } { 2 }$. If $P ( 1 , \beta ) , \beta > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is
(1) $4 x - 3 y = 2$
(2) $8 x - 2 y = 5$
(3) $7 x - 4 y = 1$
(4) $4 x - 2 y = 1$

Let $L _ { 1 }$ be a tangent to the parabola $y ^ { 2 } = 4 ( x + 1 )$ and $L _ { 2 }$ be a tangent to the parabola $y ^ { 2 } = 8 ( x + 2 )$ such that $L _ { 1 }$ and $L _ { 2 }$ intersect at right angles. Then $L _ { 1 }$ and $L _ { 2 }$ meet on the straight line:
(1) $x + 3 = 0$
(2) $2 x + 1 = 0$
(3) $x + 2 = 0$
(4) $x + 2 y = 0$

Let a line $L : 2 x + y = k , k > 0$ be a tangent to the hyperbola $x ^ { 2 } - y ^ { 2 } = 3$. If $L$ is also a tangent to the parabola $y ^ { 2 } = \alpha x$, then $\alpha$ is equal to:
(1) 12
(2) - 12
(3) 24
(4) - 24

Let $P ( a \sec \theta , b \tan \theta )$ and $Q ( a \sec \phi , b \tan \phi )$ where $\theta + \phi = \frac { \pi } { 2 }$, be two points on the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$. If the ordinate of the point of intersection of normals at $P$ and $Q$ is $- k \left( \frac { a ^ { 2 } + b ^ { 2 } } { 2 b } \right)$, then $k$ is equal to

Let $\lambda x - 2 y = \mu$ be a tangent to the hyperbola $a ^ { 2 } x ^ { 2 } - y ^ { 2 } = b ^ { 2 }$. Then $\left( \frac { \lambda } { a } \right) ^ { 2 } - \left( \frac { \mu } { b } \right) ^ { 2 }$ is equal to
(1) - 2
(2) - 4
(3) 2
(4) 4

If $m$ is the slope of a common tangent to the curves $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$ and $x ^ { 2 } + y ^ { 2 } = 12$, then $12 \mathrm {~m} ^ { 2 }$ is equal to
(1) 6
(2) 9
(3) 10
(4) 12

The normal to the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { 9 } = 1$ at the point $( 8,3 \sqrt { 3 } )$ on it passes through the point
(1) $( 15 , - 2 \sqrt { 3 } )$
(2) $( 9,2 \sqrt { 3 } )$
(3) $( - 1,9 \sqrt { 3 } )$
(4) $( - 1,6 \sqrt { 3 } )$

Let the eccentricity of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$, then the value of $c^2$ is equal to
(1) 18
(2) 20
(3) 24
(4) 32

The acute angle between the pair of tangents drawn to the ellipse $2 x ^ { 2 } + 3 y ^ { 2 } = 5$ from the point $(1, 3)$ is
(1) $\tan ^ { - 1 } \frac { 16 } { 7 \sqrt { 5 } }$
(2) $\tan ^ { - 1 } \frac { 24 } { 7 \sqrt { 5 } }$
(3) $\tan ^ { - 1 } \frac { 32 } { 7 \sqrt { 5 } }$
(4) $\tan ^ { - 1 } \frac { 3 + 8 \sqrt { 5 } } { 35 }$

Let the common tangents to the curves $4 \left( x ^ { 2 } + y ^ { 2 } \right) = 9$ and $y ^ { 2 } = 4 x$ intersect at the point $Q$. Let an ellipse, centered at the origin $O$, has lengths of semi-minor and semi-major axes equal to $OQ$ and 6, respectively. If $e$ and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac { l } { e ^ { 2 } }$ is equal to $\_\_\_\_$.

Let the tangent drawn to the parabola $y ^ { 2 } = 24 x$ at the point $( \alpha , \beta )$ is perpendicular to the line $2 x + 2 y = 5$. Then the normal to the hyperbola $\frac { x ^ { 2 } } { \alpha ^ { 2 } } - \frac { y ^ { 2 } } { \beta ^ { 2 } } = 1$ at the point $( \alpha + 4 , \beta + 4 )$ does NOT pass through the point:
(1) $( 25,10 )$
(2) $( 20,12 )$
(3) $( 30,8 )$
(4) $( 15,13 )$

A common tangent $T$ to the curves $C _ { 1 } : \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 9 } = 1$ and $C _ { 2 } : \frac { x ^ { 2 } } { 42 } - \frac { y ^ { 2 } } { 143 } = 1$ does not pass through the fourth quadrant. If $T$ touches $C _ { 1 }$ at $\left( x _ { 1 } , y _ { 1 } \right)$ and $C _ { 2 }$ at $\left( x _ { 2 } , y _ { 2 } \right)$, then $\left| 2 x _ { 1 } + x _ { 2 } \right|$ is equal to $\_\_\_\_$ .

The vertices of a hyperbola H are $( \pm 6,0 )$ and its eccentricity is $\frac { \sqrt { 5 } } { 2 }$. Let N be the normal to H at a point in the first quadrant and parallel to the line $\sqrt { 2 } x + y = 2 \sqrt { 2 }$. If $d$ is the length of the line segment of N between H and the $y$-axis then $d ^ { 2 }$ is equal to $\_\_\_\_$ .

Q84. Let a conic $C$ pass through the point $( 4 , - 2 )$ and $P ( x , y ) , x \geq 3$, be any point on $C$. Let the slope of the line touching the conic $C$ only at a single point $P$ be half the slope of the line joining the points $P$ and $( 3 , - 5 )$. If the focal distance of the point $( 7,1 )$ on $C$ is $d$, then $12 d$ equals $\_\_\_\_$

Q85. Let $L _ { 1 } , L _ { 2 }$ be the lines passing through the point $P ( 0,1 )$ and touching the parabola $9 x ^ { 2 } + 12 x + 18 y - 14 = 0$. Let $Q$ and $R$ be the points on the lines $L _ { 1 }$ and $L _ { 2 }$ such that the $\triangle P Q R$ is an isosceles triangle with base $Q R$. If the slopes of the lines $Q R$ are $m _ { 1 }$ and $m _ { 2 }$, then $16 \left( m _ { 1 } ^ { 2 } + m _ { 2 } ^ { 2 } \right)$ is equal to $\_\_\_\_$

6 Go to the solutions page
Answer the following questions.

1. [(1)] Let $A$, $\alpha$ be real numbers. Consider the equation in $\theta$: $A\sin 2\theta - \sin(\theta + \alpha) = 0$. When $A > 1$, show that this equation has at least 4 solutions in the range $0 \leq \theta < 2\pi$.
   
2. [(2)] Consider the ellipse $C: \dfrac{x^2}{2} + y^2 = 1$ in the coordinate plane. For a real number $r$ satisfying $0 < r < 1$, let $D$ be the region represented by the inequality $2x^2 + y^2 < r^2$. Show that there exists a real number $r$ $(0 < r < 1)$ such that every point $\mathrm{P}$ in $D$ satisfies the following condition. Also, find the maximum value of such $r$.
   Condition: There are at least 4 points $\mathrm{Q}$ on $C$ such that the tangent line to $C$ at $\mathrm{Q}$ and the line $\mathrm{PQ}$ are perpendicular to each other.

%% Page 7 1 Go to problem page

(1) Consider $ax^2 + bx + c > 0 \cdots\cdots\textcircled{1}$, $bx^2 + cx + a > 0 \cdots\cdots\textcircled{2}$, $cx^2 + ax + b > 0 \cdots\cdots\textcircled{3}$. The solution of this system of inequalities is $x > p$ by assumption. Here, suppose $a < 0$. Then,
$$\lim_{x \to \infty}(ax^2 + bx + c) = \lim_{x \to \infty} x^2\!\left(a + \frac{b}{x} + \frac{c}{x^2}\right) = -\infty$$
From this, for sufficiently large $x$, we have $ax^2 + bx + c < 0$, so the solution of \textcircled{1} cannot contain $x > p$. Therefore, $a \geqq 0$.
Similarly, from \textcircled{2} we get $b \geqq 0$, and from \textcircled{3} we get $c \geqq 0$.

(2) Suppose $a > 0$, $b > 0$, and $c > 0$. Then, in the same way as (1), $$\lim_{x \to -\infty}(ax^2 + bx + c) = \infty, \quad \lim_{x \to -\infty}(bx^2 + cx + a) = \infty, \quad \lim_{x \to -\infty}(cx^2 + ax + b) = \infty$$
From this, for sufficiently small $x$, all of \textcircled{1}\textcircled{2}\textcircled{3} hold. But then the solution of the system of inequalities \textcircled{1}\textcircled{2}\textcircled{3} being $x > p$ is contradicted, so from the conclusion of (1), at least one of $a$, $b$, $c$ is $0$.

(3) From (1)(2), first consider the case $a = 0$, $b \geqq 0$, $c \geqq 0$. From \textcircled{1}\textcircled{2}\textcircled{3},
$$bx + c > 0 \cdots\cdots\cdots\textcircled{4}, \quad bx^2 + cx > 0 \cdots\cdots\cdots\textcircled{5}, \quad cx^2 + b > 0 \cdots\cdots\cdots\textcircled{6}$$
From \textcircled{5}, $x(bx + c) > 0$, so from \textcircled{4}, $x > 0 \cdots\cdots\textcircled{7}$, and

(i) When $b = 0$:
From \textcircled{4}, $c > 0$, so from \textcircled{6}, $cx^2 > 0$, that is, $x \neq 0$. From this, the solution of the system of inequalities \textcircled{4}\textcircled{6}\textcircled{7} is $x > 0$.

(ii) When $b > 0$:
Since $c \geqq 0$, under \textcircled{7}, \textcircled{4} and \textcircled{6} hold, so the solution of the system of inequalities \textcircled{4}\textcircled{6}\textcircled{7} is $x > 0$.

From (i)(ii), the solution of the system of inequalities \textcircled{4}\textcircled{5}\textcircled{6} is $x > 0$.
Also, considering similarly the cases $a \geqq 0$, $b = 0$, $c \geqq 0$ and $a \geqq 0$, $b \geqq 0$, $c = 0$, the solution of the system of inequalities \textcircled{1}\textcircled{2}\textcircled{3} is $x > 0$ in all cases, so $p = 0$.

[Commentary]
This is a proof problem on systems of inequalities. Each part is considered by associating graphs.
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