Let the ellipse E:x^2 + 9y^2 = 9 intersect the positive x and y-axes at the points A and B…

Question

Let the ellipse $E:{x^2} + 9{y^2} = 9$ intersect the positive x and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $${m \over n}$$, where m and n are coprime, then $m - n$ is equal to :

Accepted Answer

The given equation of the ellipse is

$$
\begin{aligned}
& x^2+9 y^2=9 ~..........(i)\\
& \Rightarrow  \frac{x^2}{9}+\frac{y^2}{1}=1
\end{aligned}
$$

Now, equation of line $A B$ is

$x+3 y=3
~$...........(ii)

Now, point of intersection of line $x+3 y=3$ and circle

$$
\[\begin{gathered} x^2+y^2=9 \ \therefore  (3-3 y)^2+y^2=9 \ \Rightarrow  9-18 y+9 y^2+y^2=9 \ \Rightarrow  -18 y+10 y^2=0 \ \Rightarrow  18 y=10 y^2 \ \Rightarrow  y=0, \frac{9}{5} \end{gathered}$$
\]

Now, from figure, we clearly see that vertices $A, P$ and $O$ makes a $\triangle A P O$,

whose area is $\frac{1}{2} \times$ Base $\times$ Height

$$
\begin{aligned}
& \text { Area }=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{m}{n} ~~~~[Given]\\
& \therefore m=27, n=10 \\
& \Rightarrow m-n=27-10=17
\end{aligned}
$$

Solutions · 1