Question

Line through $P(a, 2)$ meets the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ at A and D and meets the coordinate axes at B and C so that PA, PB, PC, PD are in G.P., then possible values of a can be:

• A. 6
• B. -6
• C. 13/2
• D. -13/2

Answer 1

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

a can be any real number with magnitude greater than or equal to 6.

Answer 2

Let the line through $P(a, 2)$ be parameterized by $x = a + r \cos\theta$ and $y = 2 + r \sin\theta$, where $r$ is the signed distance from $P$.

1. Distances to coordinate axes:

• For point B on the x-axis ($y=0$): $0 = 2 + r_B \sin\theta \implies r_B = -2/\sin\theta$. So $PB = |r_B| = 2/|\sin\theta|$.

• For point C on the y-axis ($x=0$): $0 = a + r_C \cos\theta \implies r_C = -a/\cos\theta$. So $PC = |r_C| = |a|/|\cos\theta|$.

2. Distances to the ellipse:

Substitute $x = a + r \cos\theta$ and $y = 2 + r \sin\theta$ into the ellipse equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$:

$4(a + r \cos\theta)^2 + 9(2 + r \sin\theta)^2 = 36$ $4(a^2 + 2ar\cos\theta + r^2\cos^2\theta) + 9(4 + 4r\sin\theta + r^2\sin^2\theta) = 36$ $4a^2 + 8ar\cos\theta + 4r^2\cos^2\theta + 36 + 36r\sin\theta + 9r^2\sin^2\theta = 36$ $r^2(4\cos^2\theta + 9\sin^2\theta) + r(8a\cos\theta + 36\sin\theta) + 4a^2 = 0$

Let $r_A$ and $r_D$ be the roots of this quadratic equation. These are the signed distances from $P$ to $A$ and $D$. The product of the distances $PA \cdot PD = |r_A r_D| = \left|\frac{4a^2}{4\cos^2\theta + 9\sin^2\theta}\right| = \frac{4a^2}{4\cos^2\theta + 9\sin^2\theta}$.

3. G.P. condition:

If PA, PB, PC, PD are in G.P., then $PA \cdot PD = PB \cdot PC$.

Substitute the expressions for the products: $\frac{4a^2}{4\cos^2\theta + 9\sin^2\theta} = \left(\frac{2}{|\sin\theta|}\right) \left(\frac{|a|}{|\cos\theta|}\right)$

If $a=0$, then $P=(0,2)$. This point lies on the ellipse, so $PA=0$. For the distances to be in G.P., all terms must be zero. This would require $PB=0$, which means $P$ is on the x-axis, which is false for $P(0,2)$. Thus $a \ne 0$.

Since $a \ne 0$, we can divide by $|a|$: $\frac{4|a|}{4\cos^2\theta + 9\sin^2\theta} = \frac{2}{|\sin\theta\cos\theta|}$ $2|a| |\sin\theta\cos\theta| = 4\cos^2\theta + 9\sin^2\theta$ $|a| |2\sin\theta\cos\theta| = 4\cos^2\theta + 9\sin^2\theta$ $|a| |\sin(2\theta)| = 4\cos^2\theta + 9\sin^2\theta$

We can rewrite $4\cos^2\theta + 9\sin^2\theta = 4(1-\sin^2\theta) + 9\sin^2\theta = 4 + 5\sin^2\theta$. So, $|a| |\sin(2\theta)| = 4 + 5\sin^2\theta$.

$|a| = \frac{4 + 5\sin^2\theta}{|\sin(2\theta)|}$

To find the possible values of $|a|$, we need to find the range of the right-hand side.

Let $s = \sin^2\theta$. Then $|\sin(2\theta)| = |2\sin\theta\cos\theta| = |2\sin\theta\sqrt{1-\sin^2\theta}| = 2\sqrt{s(1-s)}$. $|a| = \frac{4+5s}{2\sqrt{s(1-s)}}$

Square both sides: $a^2 = \frac{(4+5s)^2}{4s(1-s)}$ $4a^2s(1-s) = (4+5s)^2$ $4a^2s - 4a^2s^2 = 16 + 40s + 25s^2$ $s^2(25+4a^2) + s(40-4a^2) + 16 = 0$

For $s$ to be a real value, the discriminant must be non-negative: $D = (40-4a^2)^2 - 4(25+4a^2)(16) \ge 0$ $(10-a^2)^2 - 4(25+4a^2) \ge 0$ $100 - 20a^2 + a^4 - 100 - 16a^2 \ge 0$ $a^4 - 36a^2 \ge 0$ $a^2(a^2 - 36) \ge 0$

Since $a \ne 0$, $a^2 > 0$. Therefore, we must have $a^2 - 36 \ge 0$, which implies $a^2 \ge 36$. So, $|a| \ge 6$.

Any value of $a$ such that $|a| \ge 6$ is a possible value.