The line through \(P(a, 2)\), where \(a \neq 0\), making an angle \(45^{\circ}\) with the positive direction…
The line through \(P(a, 2)\), where \(a \neq 0\), making an angle \(45^{\circ}\) with the positive direction of the \(X\)-axis meets the curve \(\frac{x^2}{9}+\frac{y^2}{4}=1\) at \(A\) and \(D\) and the coordinate axes at \(B\) and \(C\). If \(P A, P B, P C\) and \(P D\) are in a geometric progression, then \(2 a=\)
13
7
1
-13
Solution
Since equation of line passes through \(P(a, 2)\), \(a \neq 0\) making an angle \(45^{\circ}\) with positive direction of the \(X\)-axis is :
\(\frac{x-a}{\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}=r \Rightarrow x=a+\frac{r}{\sqrt{2}} \text { and } y=2+\frac{r}{\sqrt{2}}\)
So for point \(B\) (as it is on \(X\)-axis), so, \(r=-2 \sqrt{2}\) so \(B(a-2,0)\) for point \(C\) (as it is on \(Y\)-axis), so
\(r=-a \sqrt{2}\)
so, \(C(0,2-a)\)
\(\begin{aligned}
\therefore \quad P B & =\sqrt{4+4}=2 \sqrt{2} \\
P C & =\sqrt{a^2+a^2}=a \sqrt{2}
\end{aligned}\)
For points \(A\) and \(D\)
\(\begin{gathered}
\frac{\left(a+\frac{r}{\sqrt{2}}\right)^2}{9}+\frac{\left(2+\frac{r}{\sqrt{2}}\right)^2}{4}=1 \\
\Rightarrow \quad 4\left(a+\frac{r}{\sqrt{2}}\right)^2+9\left(2+\frac{r}{\sqrt{2}}\right)^2=36 \\
\Rightarrow 13\left(\frac{r^2}{2}\right)+\left(\frac{8 a}{\sqrt{2}}+\frac{36}{\sqrt{2}}\right) r+4 a^2=0 \text {, let having roots } \\
r_1=P A \text { and } r_2=P D, \text { so } \\
r_1 r_2=\frac{4 a^2}{\frac{13}{2}} \\
=\frac{8 a^2}{13}=(P A)(P D)
\end{gathered}\)
\(\because P A, P B, P C\) and \(P D\) are in GP. so
\((P A)(P D)=\left(P B^{\prime}\right)(P C) \Rightarrow \frac{8 a^2}{13}=4 a \Rightarrow 2 a=13\)
Hence, option (1) is correct.