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51.

If and k are segments of a focal chord of the parabola `y^(2)= 4ax`, then k =A. ` ( ab ) /( a- b ) `B. ` ( a ) /( b - a ) `C. ` ( b ) /( b - a ) `D. ` ( ab ) /( b - a ) `

Answer» Correct Answer - D
For latusrectum PQ,
` y ( t _ 1 + t _ 2 ) - 2x - 2a t _ 1 t _ 2 = 0 `
and ` t _ 1 t _ 2 = - 1 `
For any point ` P ( x , y ) `, then focal distance is ` a + x `.
` therefore b = a + x = a + a t _ 1 ^ 2 = a ( 1 + t _ 1 ^ 2 ) ` ...(i )
` c = a + x = a + at _ 2 ^ 2 = a + ( a ) /( t _ 1 ^ 2 ) ( because t _ 1 t _ 2 = - 1 ) `
` = ( a ( 1 + t _ 1 ^ 2 ) ) /( t _ 1^ 2) " " `... (ii)
` therefore (b )/ (c ) = t _ 1 ^ 2 `
` therefore ` From Eq. (i),
` b = a ( 1 + (b)/ (c )) `
` rArr b = a + ( ab ) / ( c ) `
` therefore c = ( ab ) /( b - a ) `
Discussion
52.

The line `2x+y+lamda=0` is a normal to the parabola `y^(2)=-8x,` is `lamda`=A. ` - 24 `B. `8 `C. `-16 `D. `24 `

Answer» Correct Answer - A
Given , parabola , `y^ 2 = 8x `
` rArr 2y ( dy )/ ( dx ) = 8 `
` rArr (dy ) / ( dx ) = ( 4 )/( y ) `
Slope of normal of parabola = ` - ( y )/ ( 4 ) ` … (i)
Given ` 2x + y + lamda = 0 ` is a equation of normal of the parabola ` y ^ 2 = 8 x `.
` therefore ` Slope of normal ` = - 2 " " `... (ii)
From Eqs. (i) and (ii) , we get
` - ( y ) /(4 ) = - 2 rArr y = 8 `
` therefore (8) ^ 2 = 8 x rArr x = 8 `
Now, putting the values of x and y in the equation of normal
` 2 ( 8 ) + 8 + lamda =0 `
` rArr lamda = -24 `
Discussion
53.

The line `lx+my+n=0` is a normal to the parabola `y^2 = 4ax` ifA. ` (( -n ) /(l ) , ( - 2am ) /( l)) `B. ` ( ( -n ) /(l) , ( 2am )/( l) ) `C. ` ( ( n ) / ( l) , ( - 2am )/ ( l)) `D. ` (( n ) /( l) , ( 2am ) /( l)) `

Answer» Correct Answer - C
Tangent of the parabola ` y^ 2 = 4ax ` at point ` P ( x_ 1, y _ 1 ) ` is
` yy _ 1 = 2a ( x +x _ 1 ) `
`rArr 2a x _ 1 - yy _ 1 + 2ax = 0 " " ` … (i)
Which is also equation of the polar of the parabola `y ^ 2 = 4ax ` . Same as the line
` lx + my + n = 0 " " `... (ii)
On comparing both lines, we get
` ( 2a)/ (l ) = ( - y ) /( m) = ( 2ax ) /( n ) `
Taking first two parts
` ( y = - ( 2am ) / ( l)) `
Taking first and last parts,
` ( x = ( n ) / ( l) ) `
` therefore ` Required pole of the line is ` { ( n ) / (l) , ( - 2am ) /(l) } `.
Discussion
54.

15 coins are tossed , the probability of getting heads will beA. ` (511 ) /( 32768) `B. ` ( 1001 ) /( 32768 ) `C. ` ( 3003 ) /( 32768 ) `D. ` ( 3005 )/( 32768 ) `

Answer» Correct Answer - C
` therefore ` Required probability ` = ""^( 15)C _ ( 10 ) (( 1 ) /( 2 ) ) ^( 10) ( ( 1)/ ( 2 ) ) ^( 5 ) `
` = ""^( 15 ) C _ 5 ( 1 ) /( 2 ^ ( 15 ) ) `
` = ( 15 xx 14 xx 13 xx 12 xx 11 ) /( 5 xx 4 xx 3xx 2 xx 1 ) xx ( 1 ) /( 2 ^ ( 15 ) ) `
` = ( 3003 )/( 32768 ) `
Discussion
55.

The odds against solving a problem by A and B are 3 :2 and 2 : 1 respectively, then the probability that the problem will be solved, isA. ` ( 3 ) /( 5 ) `B. ` ( 2 ) /( 15 ) `C. ` ( 2 ) /( 5 ) `D. ` ( 11 ) /( 5 ) `

Answer» Correct Answer - A
Given ` P ( A) = ( 3 ) /( 2 + 3 ) = ( 3 ) / ( 5 ) , P (B) = ( 2 )/ ( 2 + 1 ) = (2 ) / (3 ) `
and ` P ( barA ) = 1 - ( 3 ) /( 5 ) = ( 2 ) /( 5 ) , P ( bar B ) = 1 - ( 2 )/ (3 ) = (1 ) / ( 3 ) `
` therefore ` Required probability
` = P ( A nn bar B ) + P ( barA nn B ) + P ( bar A nn bar B ) `
` = P (A) * P (barB) + P ( barA ) * P (B) + P (bar A ) * P (barB) `
` = ( 3 )/( 5 ) * ( 1) /( 3 ) + ( 2 ) /( 5 ) * ( 2 ) /(3 ) + ( 2 )/( 5 ) * (1)/( 3 ) `
` = ( 1 )/( 5 ) + ( 4 ) /( 15 ) + ( 2 )/( 15 ) `
`= ( 3 + 4 + 2 ) /( 15 ) = ( 9 ) /( 15 ) = ( 3 )/( 5 ) `
Discussion