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A particle just clears a wall of heig...

A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)b/(a c)` (2) `"45"^o` (3) `tan^(-1)(b c)/(a(c-a)` (4) `tan^(-1)(b c)/a`

A

`(3,0) and (0,2)`

B

`(-8//5,2sqrt(161)//15) and (-9//5,8//5)`

C

`(8//5,2sqrt(161)//15 and (0,2)`

D

`(3,0) and (-9//5,8//5)`

Text Solution

Verified by Experts

Given ellipse `(x^(2))/(9)+(y^(2))/(4)=1`
The equation of tangent having slope m is
`y=mx+sqrt(9m^(2)+4)`
Sqaring , we have
`16+9m^(2)-24m=9m^(2)+4`
or `m=(12)/(24)=(1)/(2)`
Threrefpre, the equation of tangent is `y-4=-(x-3)pr x-2y+5=0 " "(1)`
Let the point of contact on the curve be `B(alpha,beta)`
Equation of tangent at this point is `(xalpha)/(9)+(ybeta)/(4)-1=0" "(2)`
Comparing equations (1) and (2)
`(alpha//9)/(1)=(beta//4)/(-2)=-(1)/(5)`
or `alpha=-(9)/(5),beta=(8)/(5)" " :. B-=(-(9)/(5),(8)/(5))`
Another slope of tangent is `oo` then the equation of tanents is `x=3` and the coresponding point of cocntact is A(3,0)
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