a line passes through the point P (2,3) and makes an angle \[\theta\] with positive direction of x - axis. If it meets the lines represented by \[x^2 - 2xy - y^2 = 0\] at the points A and B. If PA.PB = 17, then find the value of \[\frac{10}{\pi} * \theta\]
@eliassaab
@mukushla @waterineyes
@apoorvk ?
@vamgadu
can someone help me
its urgent for me.......
@mathslover
\[\left(x-y\right)^2 + \left(\sqrt2y\right)^2=0\]
negative sign in between not positive.
ok
Equation of line must be \(y = mx + 3 -2m\) where \(m=\tan \theta\). |dw:1342782842699:dw|
I am sorry I made a foolish mistake in representation both the lines are supposed to have positive slope, I hope you can correct it yourself.
Ok then what do I have to do..............
Point A could be represented as. (2 + PAcos theta, 3 + PA sin theta). Point B, (2 + PB cos theta, 3 + PB sin theta). Point A and B must be lie on the line y=mx+3−2m
Also, PA.PB = 17
yes
Now solve it yourself. I have give you enough hints.
given*
So I can apply distance formula
on PA and PB and then equate (PA.PB)^2 = 17^2
Why would you apply distance formula? PA and PB are already distances.
No I mean distance between P and A = PA distance between P and B = PB
It won't work.
y ? then what should I do........
@amistre64 @eliassaab @UnkleRhaukus @mukushla someone please help me................
@experimentX
what bit are you up to?
@UnkleRhaukus I tried finding the value of PA and PB using distance formula and then equating PA.PB = 17 but that didn't want...........
you do have a drawing ?
x^2 -2xy -y^2=0 on solving treating it as quad eqn in x, 2 lines are : x = y (1+ sqrt(2)) and x= y(1- sqrt(2)) our line is y= (tan@) x +c it passes through (2,3) => c=3-2tan@ =>eqn is y=xtan@ + 3-2tan@ A = intersection of y=xtan@ + 3-2tan@ and x = y (1+ sqrt(2)) solve for x and y similarly find B then PA and PB =>find tan@ or @ you have the ans then my method seems very long and monotonous..hmm..
@shubhamsrg thanks a lot for ur help I did find my answer using the same method but couldn't find an easier one......... Can u suggest an easier method ?
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