Question
A normal is drawn at the point P to the parabola y^2=8 x, which is inclined at 60^ with the straight line y=8. Then the point P lies on the straight line
BITSAT
Mathematics
Parabola
2022
BITSAT 2022
BITSAT Mathematics Question (2022) — Solution
Options
- A. 2 x+y-12-4 3 =0
- B. 2 x-y-12+4 3 =0
- C. 2 x-y-12-4 3 =0
- D. None of these
Answer
C. 2 x-y-12-4 3 =0
Step-by-step solution
For the parabola y^2=4 a x, the equation of normal at P (a m^2,-2 a m ) is y=m x-2 a m-a m^3. Here, m= 60^ = 3 P (a( 3 )^2,-2 a( 3 ) ) (6,-4 3 ) ( a=2) Thus, P satisfies 2 x-y-12-4 3 =0
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