JEE Main 202120 JulyMorning Shift
- Exam
- JEE Main
- Year
- 2021
- Date
- 20 July
- Shift
- Morning Shift
- Subject
- Mathematics
- Chapter
- Parabola
JEE Main 2021 · 20 July · Morning Shift · Mathematics · Parabola
Let y = mx + c, m > 0 be the focal chord of y ^ 2 = - 64x , which...
Let be the focal chord of , which is tangent to . Then, the value of is equal to______
Correct answer
0
Step-by-step solution
We have,
Coordinates of focus of parabola is : \left(- 16, 0\right)y = mx + c
is tangent to , thereforey = m \left(x + 10\right)
Then,4\sqrt{2}\left(m + c\right) = 4\sqrt{2}\left(\dfrac{17}{2\sqrt{2}}\right) = 34