Question

Consider a conic $C: y^2 = 4x$. Let $\triangle PQR$ be an equilateral triangle with side length $k$ where $P$ be any point on $C$, $Q$ be the foot of perpendicular from $P$ upon the directrix of $C$ and $R$ be the focus of $C$. A circle $C_2$ is inscribed in another conic $C_1: y^2 = k(x+1)$ which touches $C_1$ at the points where $C_1$ cuts the $y$-axis. $C_3$ is an ellipse whose auxiliary circle is $C_2$ and major axis coincides with the axis of symmetry of $C_1$ and whose length of minor axis is 4.

Identify the correct statement(s) for $C_3$.

• A. Eccentricity is $\frac{1}{2}$.
• B. Focal length is 4.
• C. Length of latus-rectum is $2\sqrt{2}$.
• D. Director circle is $x^2 + y^2 - 4x - 8 = 0$.

Answer 1

Answer: (B), (C), (D)

Focal length is 4., Length of latus-rectum is $2\sqrt{2}$., Director circle is $x^2 + y^2 - 4x - 8 = 0$.

Answer 2

The given conic is $C: y^2 = 4x$. This is a parabola with $a=1$. The focus of $C$ is $R = (1, 0)$. The directrix of $C$ is $x = -1$. Let $P(x_P, y_P)$ be a point on $C$, so $y_P^2 = 4x_P$. $Q$ is the foot of the perpendicular from $P$ upon the directrix $x = -1$, so $Q = (-1, y_P)$. $\triangle PQR$ is an equilateral triangle with side length $k$. The distance $PQ = \sqrt{(x_P - (-1))^2 + (y_P - y_P)^2} = \sqrt{(x_P + 1)^2} = |x_P + 1|$. Since $x_P \ge 0$ for $y_P^2 = 4x_P$, $PQ = x_P + 1$. The distance $QR = \sqrt{(-1 - 1)^2 + (y_P - 0)^2} = \sqrt{(-2)^2 + y_P^2} = \sqrt{4 + y_P^2}$. Since $\triangle PQR$ is equilateral, $PQ = QR = k$. So $k = x_P + 1$ and $k^2 = 4 + y_P^2$. From $k = x_P + 1$, we get $x_P = k - 1$. Substitute $x_P$ into $y_P^2 = 4x_P$: $y_P^2 = 4(k - 1)$. Substitute $y_P^2$ into $k^2 = 4 + y_P^2$: $k^2 = 4 + 4(k - 1) = 4 + 4k - 4 = 4k$. Since $k$ is a side length, $k > 0$. Dividing by $k$, we get $k = 4$.

The conic $C_1$ is $y^2 = k(x+1)$, which is $y^2 = 4(x+1)$. This is a parabola with vertex $(-1, 0)$ and axis of symmetry $y=0$ (the x-axis). $C_1$ cuts the y-axis when $x=0$: $y^2 = 4(0+1) = 4$, so $y = \pm 2$. The intersection points are $(0, 2)$ and $(0, -2)$. A circle $C_2$ is inscribed in $C_1$ and touches $C_1$ at $(0, 2)$ and $(0, -2)$. Since the points of tangency are symmetric about the x-axis, the center of $C_2$ lies on the x-axis. Let the center be $(h_c, 0)$. The equation of $C_2$ is $(x - h_c)^2 + y^2 = r^2$. The points $(0, 2)$ and $(0, -2)$ lie on $C_2$: $(0 - h_c)^2 + 2^2 = r^2 \implies h_c^2 + 4 = r^2$. The radius to the point of tangency is perpendicular to the tangent. The tangent to $y^2 = 4(x+1)$ at $(0, 2)$ has slope $\frac{dy}{dx}$ from $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$. At $(0, 2)$, the slope of the tangent is $m_t = \frac{2}{2} = 1$. The slope of the radius from $(h_c, 0)$ to $(0, 2)$ is $m_r = \frac{2 - 0}{0 - h_c} = -\frac{2}{h_c}$. Since $m_r m_t = -1$, we have $(-\frac{2}{h_c})(1) = -1 \implies h_c = 2$. The center of $C_2$ is $(2, 0)$. The radius squared is $r^2 = h_c^2 + 4 = 2^2 + 4 = 8$. The equation of $C_2$ is $(x - 2)^2 + y^2 = 8$.

$C_3$ is an ellipse whose auxiliary circle is $C_2$ and major axis coincides with the axis of symmetry of $C_1$. The axis of symmetry of $C_1$ is the x-axis ($y=0$). The major axis of $C_3$ is along the x-axis. The auxiliary circle $C_2$ is centered at $(2, 0)$, so the center of the ellipse $C_3$ is also $(2, 0)$. The equation of the auxiliary circle of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ with $a>b$ is $(x - h)^2 + (y - k)^2 = a^2$. Comparing $(x - 2)^2 + y^2 = 8$ with $(x - 2)^2 + y^2 = a_{ell}^2$, we get $a_{ell}^2 = 8$. Since the major axis is along the x-axis, $a_{ell}^2$ is the larger denominator. $a_{ell} = \sqrt{8} = 2\sqrt{2}$. The length of the minor axis is given as 4, so $2b_{ell} = 4 \implies b_{ell} = 2$. $b_{ell}^2 = 4$. The equation of the ellipse $C_3$ is $\frac{(x - 2)^2}{8} + \frac{y^2}{4} = 1$.

Now let's check the statements for $C_3$:

1. Eccentricity $e$: $b_{ell}^2 = a_{ell}^2 (1 - e^2) \implies 4 = 8(1 - e^2) \implies \frac{1}{2} = 1 - e^2 \implies e^2 = \frac{1}{2} \implies e = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. Statement says $e = \frac{1}{2}$. Incorrect.
2. Focal length: The distance from the center $(2, 0)$ to each focus is $a_{ell}e = (2\sqrt{2})(\frac{1}{\sqrt{2}}) = 2$. The foci are at $(2 \pm 2, 0)$, which are $(4, 0)$ and $(0, 0)$. The focal length (distance between foci) is $\sqrt{(4-0)^2 + (0-0)^2} = 4$. Statement says focal length is 4. Correct.
3. Length of latus-rectum: $\frac{2b_{ell}^2}{a_{ell}} = \frac{2(4)}{2\sqrt{2}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$. Statement says length of latus-rectum is $2\sqrt{2}$. Correct.
4. Director circle: The equation of the director circle for $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ is $(x - h)^2 + (y - k)^2 = a^2 + b^2$. For $C_3: \frac{(x - 2)^2}{8} + \frac{y^2}{4} = 1$, the director circle is $(x - 2)^2 + y^2 = 8 + 4 = 12$. $(x - 2)^2 + y^2 = 12$ $x^2 - 4x + 4 + y^2 = 12$ $x^2 + y^2 - 4x - 8 = 0$. Statement says director circle is $x^2 + y^2 - 4x - 8 = 0$. Correct.

The correct statements are 2, 3, and 4.