# 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?

68 Points
13 years ago

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

All the best.

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Karanveer
33 Points
4 years ago

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 contact
Since 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/6
Hope it helps.