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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? 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?
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 Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Ajit Singh Verma IITD
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
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Askiitians Expert
Ajit Singh Verma IITD
Answer is π/6 Step by Step Explanation :Equation of given hyperbola can be written as SoSimilarly General point on ellipse Using T=0 on ellipse We getWhere (p,q) is point of contactSince it's equation of tangent, so by comparing this with given equation of tangentWe get p= And q = Since p = So π/6Hope it helps.
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