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        a rectangular billiard table has vertices at  p(0,0), q(0,7) ,r(10,7)&s(10,0) . a small billiard ball starts at m(3,4) & moves in astraight line to the top of the table, bounces to the right of the table, then comes to rest at n(7,1) .what is the y-cordinate of the point where ball hits the right side of table?
8 years ago

## Answers : (2)

68 Points
							Dear ffggfg hkjj,

I guess the above figure is what is meant by the question, We will apply the laws of reflection at the strike points of the ball. So the angle at which the ball is reflected after the first strike is equal to angle at which it had striked i.e A .
slopes:
PM : tan(90-A) = 3/x-3   or cot A =  3/x-3  -----------------------1
PQ: tan(-(90-A)) = y-7/10-x  or -Cot A = y-7/10-x-------------------2
QN: tan(90-A) = y-1/3 or cot A = y-1/3----------------------------3
divide 1 by 2 and 1 by 3
we get
10x-30 = xy-3y +21
x+ 9  = xy-3y +3
solving we get
x= 19/3 and y = 37/10

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Ajit Singh Verma IITD

8 years ago
Karanveer
33 Points
							Answer is π/6 Step by Step Explanation :Equation of given hyperbola can be written as $\frac{x^{2}}{1/4} + \frac{y^{2}}{1/9} = 1$So$a^{2} = 1/4$$a= 1/2$Similarly $b= 1/3$General point on ellipse $(1/2 \sec \theta,1/3 \tan \theta)$Using T=0 on ellipse We get$4px-9qy = 1$Where (p,q) is point of contactSince it's equation of tangent, so by comparing this with given equation of tangent$4x-3y = \sqrt{3}$We get p= $1/\sqrt{3}$ And q = $1/3\sqrt{3}$Since p = $1/2\sec \theta$$1/2\sec \theta = 1/\sqrt{3}$$sec\theta = 2/\sqrt{3}$So $\Theta = %u03C0/6$ π/6Hope it helps.

3 months ago
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• Mind Map
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