If line through point $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at A and D and meets the coordinate axis at B and C

Question

If line through point $P(a,2)$ 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 geometric progression, then the possible values of $a$ can be
$(A)5\hspace{1cm}(B)8\hspace{1cm}(C)10\hspace{1cm}(D)-7$

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I could not solve this question, I inferred from question that $PA\cdot PD=PB\cdot PC$ and $PA\cdot PD=PT^2$, where $T$ is the point of tangency. But I could not solve further. This is a multiple correct choice type question. Please help me.
Thanks.

Answer 1

WLOG we can take $a>0$, $A$ nearer to $P$ than $D$ and notice that the only case we must consider is when $B$ is the intersection with $x$-axis and $C$ is the intersection with $y$-axis, for otherwise those four segments cannot form a geometric progression. If $b$ is the $x$ coordinate of $B$, the equation of line $PB$ is $y=2(x-a)/(a-b)+2$ so that the $y$ coordinate of $C$ is $y_C=-2b/(a-b)$.

Combining this equation with that of the ellipse, we can readily find the $y$ coordinate of $A$ and $D$:

$$ y_A= \frac{2 \left(3 \sqrt{a^2-2 a b+9}-a b+b^2\right)}{a^2-2 a b+b^2+9}, \quad y_D= \frac{2 \left(-3 \sqrt{a^2-2 a b+9}-a b+b^2\right)}{a^2-2 a b+b^2+9}. $$

We know that $PA:PB=PB:PC=PC:PD$ and this relation also holds for the $y$ components of the segments, that is: $$ (y_P - y_A):(y_P - y_B) = (y_P - y_B):(y_P - y_C) = (y_P - y_C):(y_P - y_D). $$ Inserting here the expressions given above for $y_C$, $y_A$, $y_D$, as well as $y_P=2$ and $y_B=0$, we can solve for $a$ and $b$. The only acceptable positive solution is: $$ a=3 \sqrt{2+\sqrt{13}}\approx 7.10281, $$ but of course the opposite value, by symmetry, is also a valid solution. As you can see, this is not far from your $(D)$ choice but it is not the same. So the exercise is wrong.

Answer 2

Consider the ellipse,
Ellipse image(Click here)

The Equation of line is, $$ y-2=(x-a)tan\theta, $$ The P(A) is, $$ P(A) = (\space a+(r)cos\theta,2+(r)sin\theta\space ), $$ where r is the distance between P(A) and P(a,2).

Solving equation of the straight line by putting x=0 and y=0,we get $$ P(B)=(\space a-2cot\theta,0\space)\space and \space P(C)=(\space 0,2-(a)tan\theta\space), $$ We also get distances, $$ PB=\sqrt(\space(2cot\theta)^2+4\space)=2cosec\theta, $$ and $$ PC=\sqrt(\space a^2+(atan\theta)^2\space)=asec\theta, $$ Similarly we get PD and PA by solving straight line and ellipse equation, Now $$ PA.PD=PB.PC,\space PB/PA = PD/PC $$ $$ 4a^2/(4 + 5sin^2\theta) = 2a/sin\theta.cos\theta, $$ $$ solving, we \space get\space a=(\space(13-5cos2\theta)/2sin2\theta\space). $$ $$ then \space we \space get,\space a=(\space (13-5(\space 1-tan^2\theta)/(1+tan^2\theta)\space)/4tan\theta/(1+tan^2\theta)\space) $$ We then get a quadratic equation of tan(theta),
Solving putting D=0 we get,
$$ a^2\space>36,\space therefore \space a\space>6 $$ The options are (B) and (C).