JEE Main 2021 — Parabola Question with Solution
From: JEE Main 2021 (Online) 20th July Morning Shift
Question
Let y = mx + c, m > 0 be the focal chord of y2 = 64x, which is tangent to (x + 10)2 + y2 = 4. Then, the value of 4 (m + c) is equal to _____________.
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Show full solutionCorrect answer: 34
Correct answer
34
Step-by-step explanation
y2 = 64x
focus : (16, 0)
y = mx + c is focal chord
c = 16 m ...........(1)
y = mx + c is tangent to (x + 10)2 + y2 = 4
y = m(x + 10) 2
c = 10m 2
16m = 10m 2
6m = 2 (m > 0)
9m2 = 1 + m2
m = & c =
= 34
focus : (16, 0)
y = mx + c is focal chord
c = 16 m ...........(1)
y = mx + c is tangent to (x + 10)2 + y2 = 4
y = m(x + 10) 2
c = 10m 2
16m = 10m 2
6m = 2 (m > 0)
9m2 = 1 + m2
m = & c =
= 34
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