# 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

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
$(A)5\hspace{1cm}(B)8\hspace{1cm}(C)10\hspace{1cm}(D)-7$

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.

• Where did you find this question? Commented Oct 8, 2015 at 16:23
• In my jee preparation book?But why are you asking so?@Aretino Commented Oct 9, 2015 at 3:56
• To be sure it is not from an ongoing contest. Commented Oct 9, 2015 at 7:02
• Do you know how is it to be solved?@Aretino Commented Oct 9, 2015 at 7:48
• How do you know that $PA\cdot PD=PT^2$? Commented Oct 9, 2015 at 8:46

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.

Consider the ellipse,
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),
$$a^2\space>36,\space therefore \space a\space>6$$ The options are (B) and (C).