please  help me  by  solving  this question. i  am  unable  to   do so...

2081 Points
2 years ago
dear muskan, the soln is insanely lengthy. so i ll only tell the steps to solve it.
first we find the coordinates of B and K.
for coordinates of B, solve T: 3y=x+10 and P: y^2 – 2y – 4x+9=0 simultaneously and u ll get B(11, 7)
at any point (x1, y1) on the parabola we have the eqn of tangent as:
yy1+9= y+y1+2(x+x1) (std result)
or y(y1 – 1)= 2x + (2x1+y1 – 9)
so, slope of tangent at any pt = 2/(y1 – 1)
since normal and tangent are at 90deg, we have
2/(y1 – 1)= – 1/slope of normal= – 3 as slope of 27y-9x+10=0 is 1/3.
so, we get K(19/9, 1/3).
now, we can write P: (y – 1)^2= 4(x – 2)
which means that its axis is y – 1= 0 or y=1.
line joining BK is y – 7= [(7 – 1/3)/(11 – 19/9)](x – 11),
find x when y=1. call this point P.
find BP and KP as alpha and beta by dist formula.
you ll see that BP= 10 and KP= 10/9.
so alpha+beta= alpha*beta= 100/9
now, given quad eqn has real and distinct roots if and only if its discriminant is strictly greater than 0.
after setting b^2 – 4ac greater than 0 and simplifying, we get
a^2 – 5a – 14 less than 0.
or (a – 7)(a+2) less than 0.
so a lies in ( – 2, 7).
since a is an integer, it can attain values – 1, 0, 1, 2, 3, 4, 5, 6
hence, the no of integral values of a is 8.
note that the calculations involved in this ques are extremely lengthy, time consuming and prone to mistakes. this kind of ques is not asked in jee. so no need to worry.
KINDLY APPROVE :))