The line through P(a, 2),where a neq 0,making | vedclass

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(A) The equation of the line passing through $P(a, 2)$ with an angle of $45^{\circ}$ to the $X$-axis is given by $\frac{x-a}{\cos 45^{\circ}} = \frac{

Vedclass · Accepted Answer

$13$

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(A) The equation of the line passing through $P(a, 2)$ with an angle of $45^{\circ}$ to the $X$-axis is given by $\frac{x-a}{\cos 45^{\circ}} = \frac{y-2}{\sin 45^{\circ}} = r$,which simplifies to $x = a + \frac{r}{\sqrt{2}}$ and $y = 2 + \frac{r}{\sqrt{2}}$.For point $B$ on the $X$-axis,$y=0 \Rightarrow 2 + \frac{r}{\sqrt{2}} = 0 \Rightarrow r = -2\sqrt{2}$. Thus $PB = |r| = 2\sqrt{2}$.For point $C$ on the $Y$-axis,$x=0 \Rightarrow a + \frac{r}{\sqrt{2}} = 0 \Rightarrow r = -a\sqrt{2}$. Thus $PC = |r| = |a|\sqrt{2}$.For points $A$ and $D$ on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,substitute $x$ and $y$:$\frac{(a + r/\sqrt{2})^2}{9} + \frac{(2 + r/\sqrt{2})^2}{4} = 1$.Expanding this,we get $4(a^2 + \sqrt{2}ar + r^2/2) + 9(4 + 2\sqrt{2}r + r^2/2) = 36$.$13r^2/2 + (4\sqrt{2}a + 18\sqrt{2})r + 4a^2 = 0$.The product of roots $PA \cdot PD = \frac{4a^2}{13/2} = \frac{8a^2}{13}$.Since $PA, PB, PC, PD$ are in $GP$,$PA \cdot PD = PB \cdot PC$.$\frac{8a^2}{13} = (2\sqrt{2})(|a|\sqrt{2}) = 4|a|$.Assuming $a > 0$,$\frac{8a^2}{13} = 4a \Rightarrow 2a = 13$.

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 $PA, PB, PC$ and $PD$ are in a geometric progression,then $2a=$

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