Question

Consider the ellipse $E: \frac{x^2}{6} + \frac{y^2}{3} = 1$. Let $E_1$ denote the locus of the points of intersection of a pair of perpendicular tangents to $E$. A vertical line passing through the point $P(h, 0)$ intersects $E$ and $E_1$ at $Q$ and $R$ respectively in the first quadrant. Let the tangent to the ellipse at point $Q$ meet x axis at the point $T$. If $\Delta(h)$ = area of the triangle $PRT$, $\Delta_1 = \underset{1\leq h\leq 2}{max}\Delta(h)$ and $\Delta_2 = \underset{1\leq h\leq 2}{min}\Delta(h)$, then $\Delta_1^2 - 4\Delta_2^2$ equals 45. Range [44.99 to 45.01]

Answer 1

45

Answer 2

The ellipse $E$ is given by $\frac{x^2}{6} + \frac{y^2}{3} = 1$. This is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2=6$ and $b^2=3$.

The locus of the points of intersection of a pair of perpendicular tangents to $E$ is the director circle $E_1$. The equation of the director circle is $x^2 + y^2 = a^2 + b^2$.

So, $E_1$ is $x^2 + y^2 = 6 + 3 = 9$.

A vertical line $x=h$ passes through $P(h, 0)$, where $1 \leq h \leq 2$. This line intersects $E$ and $E_1$ at $Q$ and $R$ respectively in the first quadrant.

For $E$: $\frac{h^2}{6} + \frac{y^2}{3} = 1 \implies \frac{y^2}{3} = 1 - \frac{h^2}{6} = \frac{6-h^2}{6} \implies y^2 = \frac{6-h^2}{2}$. Since $Q$ is in the first quadrant, $y_Q = \sqrt{\frac{6-h^2}{2}}$. So $Q = (h, \sqrt{\frac{6-h^2}{2}})$.

For $E_1$: $h^2 + y^2 = 9 \implies y^2 = 9-h^2$. Since $R$ is in the first quadrant, $y_R = \sqrt{9-h^2}$. So $R = (h, \sqrt{9-h^2})$.

For $1 \leq h \leq 2$, $h^2 \in [1, 4]$. $6-h^2 \in [2, 5]$, so $y_Q = \sqrt{\frac{6-h^2}{2}} \in [1, \sqrt{5/2}]$. $9-h^2 \in [5, 8]$, so $y_R = \sqrt{9-h^2} \in [\sqrt{5}, \sqrt{8}]$. Since $\sqrt{5} > \sqrt{5/2} \geq 1$, we have $y_R > y_Q > 0$ for $h \in [1, 2]$. The points $P, Q, R$ are collinear on the vertical line $x=h$. $P=(h,0)$, $Q=(h, y_Q)$, $R=(h, y_R)$.

The tangent to the ellipse $E$ at $Q(x_0, y_0)$ is $\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$. Here $(x_0, y_0) = (h, \sqrt{\frac{6-h^2}{2}})$, $a^2=6$, $b^2=3$. The tangent equation is $\frac{xh}{6} + \frac{y\sqrt{\frac{6-h^2}{2}}}{3} = 1$. This tangent meets the x-axis at $T$. Set $y=0$: $\frac{x_T h}{6} = 1 \implies x_T = \frac{6}{h}$. So $T = (\frac{6}{h}, 0)$.

For $1 \leq h \leq 2$, $h \in [1, 2]$, so $\frac{6}{h} \in [3, 6]$. $x_T = \frac{6}{h} \geq 3$ and $x_P = h \leq 2$. So $x_T > x_P$.

Consider the triangle $PRT$. Vertices are $P(h, 0)$, $R(h, \sqrt{9-h^2})$, $T(\frac{6}{h}, 0)$. The base $PT$ lies on the x-axis. Length of base $PT = |x_T - x_P| = |\frac{6}{h} - h|$. Since $h \in [1, 2]$, $h^2 \in [1, 4]$, $6-h^2 \in [2, 5]$. $\frac{6}{h} - h = \frac{6-h^2}{h}$. Since $h>0$ and $6-h^2 \geq 2 > 0$, $\frac{6}{h} - h > 0$. Length of base $PT = \frac{6}{h} - h$. The height of the triangle is the perpendicular distance from $R$ to the x-axis, which is $y_R = \sqrt{9-h^2}$. The area of $\Delta PRT$ is $\Delta(h) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \left(\frac{6}{h} - h\right) \sqrt{9-h^2} = \frac{1}{2} \frac{6-h^2}{h} \sqrt{9-h^2}$.

We need to find the maximum and minimum values of $\Delta(h)$ for $h \in [1, 2]$. Consider $\Delta(h)^2 = \frac{1}{4} \frac{(6-h^2)^2}{h^2} (9-h^2)$. Let $u=h^2$. Since $h \in [1, 2]$, $u \in [1, 4]$. Let $G(u) = \frac{1}{4} \frac{(6-u)^2(9-u)}{u} = \frac{1}{4u}(36-12u+u^2)(9-u) = \frac{1}{4u}(324 - 36u - 108u + 12u^2 + 9u^2 - u^3) = \frac{1}{4u}(324 - 144u + 21u^2 - u^3) = \frac{1}{4}(\frac{324}{u} - 144 + 21u - u^2)$. To find extrema, we differentiate $G(u)$ with respect to $u$: $G'(u) = \frac{1}{4}(-\frac{324}{u^2} + 21 - 2u) = \frac{1}{4u^2}(-324 + 21u^2 - 2u^3) = -\frac{1}{4u^2}(2u^3 - 21u^2 + 324)$. Let $k(u) = 2u^3 - 21u^2 + 324$. We need to check the sign of $k(u)$ for $u \in [1, 4]$. $k'(u) = 6u^2 - 42u = 6u(u-7)$. For $u \in [1, 4]$, $u>0$ and $u-7<0$, so $k'(u) < 0$. Thus, $k(u)$ is a strictly decreasing function on $[1, 4]$. $k(1) = 2(1)^3 - 21(1)^2 + 324 = 2 - 21 + 324 = 305$. $k(4) = 2(4)^3 - 21(4)^2 + 324 = 2(64) - 21(16) + 324 = 128 - 336 + 324 = 116$. Since $k(1)=305 > 0$ and $k(4)=116 > 0$, and $k(u)$ is decreasing, $k(u) > 0$ for all $u \in [1, 4]$. Therefore, $G'(u) = -\frac{1}{4u^2} k(u) < 0$ for all $u \in [1, 4]$. $G(u)$ is a strictly decreasing function on $[1, 4]$. Since $\Delta(h)^2 = G(h^2)$, $\Delta(h)^2$ is strictly decreasing for $h \in [1, 2]$. Since $\Delta(h) > 0$, $\Delta(h)$ is also strictly decreasing for $h \in [1, 2]$.

The maximum value of $\Delta(h)$ occurs at $h=1$: $\Delta_1 = \Delta(1) = \frac{1}{2} \frac{6-1^2}{1} \sqrt{9-1^2} = \frac{1}{2} \cdot 5 \cdot \sqrt{8} = \frac{5}{2} \cdot 2\sqrt{2} = 5\sqrt{2}$.

The minimum value of $\Delta(h)$ occurs at $h=2$: $\Delta_2 = \Delta(2) = \frac{1}{2} \frac{6-2^2}{2} \sqrt{9-2^2} = \frac{1}{2} \frac{6-4}{2} \sqrt{9-4} = \frac{1}{2} \frac{2}{2} \sqrt{5} = \frac{1}{2} \cdot 1 \cdot \sqrt{5} = \frac{\sqrt{5}}{2}$.

We need to calculate $\Delta_1^2 - 4\Delta_2^2$. $\Delta_1^2 = (5\sqrt{2})^2 = 25 \times 2 = 50$. $\Delta_2^2 = (\frac{\sqrt{5}}{2})^2 = \frac{5}{4}$. $4\Delta_2^2 = 4 \times \frac{5}{4} = 5$.

$\Delta_1^2 - 4\Delta_2^2 = 50 - 5 = 45$.

Explanation of the solution:

1. Identify the equation of the ellipse $E$ and its parameters $a^2$ and $b^2$.
2. Determine the equation of $E_1$, the locus of points of intersection of perpendicular tangents, which is the director circle $x^2+y^2=a^2+b^2$.
3. Find the coordinates of points $Q$ and $R$ by intersecting the vertical line $x=h$ with $E$ and $E_1$ respectively, considering the first quadrant condition.
4. Find the equation of the tangent to $E$ at point $Q$.
5. Find the coordinates of point $T$ by intersecting the tangent line with the x-axis.
6. Identify the vertices of the triangle $PRT$ and calculate its area $\Delta(h)$ using the base $PT$ on the x-axis and the height as the y-coordinate of $R$.
7. Analyze the function $\Delta(h)$ for $h \in [1, 2]$ to find its maximum $\Delta_1$ and minimum $\Delta_2$. This is done by analyzing the square of the area function, $G(u) = \Delta(h)^2$ with $u=h^2$.
8. Differentiate $G(u)$ and analyze the sign of the derivative $G'(u)$ on the interval $u \in [1, 4]$.
9. Conclude that $G(u)$ is decreasing on $[1, 4]$, which implies $\Delta(h)$ is decreasing on $[1, 2]$.
10. Determine $\Delta_1$ as $\Delta(1)$ and $\Delta_2$ as $\Delta(2)$.
11. Calculate $\Delta_1^2$ and $\Delta_2^2$.
12. Compute the final value $\Delta_1^2 - 4\Delta_2^2$.