Can someone tell me how to approach and solve this problem..?

If \[(1+ax)^{n}=1+8x+24x^2+...\] and a line through P(a , n) cuts the circle \[x^2+y^2=4\] in A and B then PA.PB is equal to \[(a)4\] \[(b)8\] \[(c)16\] \[(d)32\]

Please explain the solution in DETAIL .....

#Algebra

Easy Math Editor

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`2^{34}`	$2^{34}$
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Comments

$(1+ax)^{n}=1+n*ax+\dfrac{n(n-1)}{2}*a^{2}*x^{2}+.........=1+8x+24x^{2}+.......Comparing\ \\co-efficients,we\ get a*n=8\ and\ (n-1)*(a)=6.Solving\ these\ two\ we\ get:a=2\ and\ n=4.So\ the\ \\point\ P\ is(2,4).Now,length\ of\ the\ tangent\ on\ the\ given\ circle\ is\\=\sqrt{2^{2}+4^{2}-4}.This\ comes\ to\ 4.Now,PA*PB=PT^{2}=4^{2}.\\Therefore,PA*PB=16.PT:length\ of\ the\ tangent\ to\ the\ circle\ from\ P.$

satyendra kumar - 6 years, 7 months ago

Nice solution! I advise you not to use latex in typing text! It will look more pleasant! See this for more info!

Pranjal Jain - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$