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Standard XII

Mathematics

Distance Formula

A1, A2,....An...

Question

$A_{1}, A_{2}, . . . . A_{n}$ are points on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n O A_{n - 1}, O$ being the origin . If $O A_{1} = 1$ then the coordinates of $A_{8}$ are $(3 a \sqrt{2}, 3 a \sqrt{2})$ where $a$ is equal to

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Solution

We can see that for $n = 8$ we have
$O A_{8} = 8 O A_{7} = 8 (7 O A_{6}) = 8 \times 7 (6 A_{5}) . . . . . . . . . . = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 O A_{1}$
Hence we get the expression written below after substituting the value of $O A_{1} = 1$
$O A_{8} = 8! O A_{1} = 40320$
$(3 a \sqrt{2}, 3 a \sqrt{2}) = (O A_{8} cos 45^{o}, O A_{8} sin 45^{o})$
$a = \frac{1}{3 \sqrt{2}} \times 40320 \times \frac{1}{\sqrt{2}} = 6720$

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A1,A2,....An are points on the line y=x lying in the positive quadrant such that OAn=nO An−1,O being the origin . If OA1=1 then the coordinates of A8 are (3a√2,3a√2) where a is equal to

We can see that for n=8 we haveOA8=8OA7=8(7OA6)=8×7(6A5)..........=8×7×6×5×4×3×2×1 OA1Hence we get the expression written below after substituting the value of OA1=1 OA8=8!OA1=40320(3a√2,3a√2)=(OA8cos45o,OA8sin45o)a=13√2×40320×1√2=6720

$A_{1}, A_{2}, . . . . A_{n}$ are points on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n O A_{n - 1}, O$ being the origin . If $O A_{1} = 1$ then the coordinates of $A_{8}$ are $(3 a \sqrt{2}, 3 a \sqrt{2})$ where $a$ is equal to