Application of DerivativeHard
Question
A tangent to the curve y = x2 + 3x passes through a point (0, -9) if it is drawn at the point-
Options
A.(-3, 0)
B.(1, 4)
C.(0, 0)
D.(-4 , 4)
Solution
y = x2 + 3x passes through a point (0, -9)
Let the point is P(x1, y1) on the curve
So,
= 2x + 3
tangent
= 2x1 + 3
(y - y1) = (x - x1) (2x1 + 3)
y - y1 = 2xx1 + 3x - 2x21 - 3x1
x(2x1 + 3) - y + y1 - 2x21 - 3x1 = 0
it passes (0, 9) then
0 (2x1 + 3) + 9 + y1 - 2x21 - 3x1 = 0
(x1, y1) also lies on the curve
= y1 - 2x21 - 3x1 + 9 = 0 ..... (1)
so y1 = x21 + 3x1 ..... (2)
From Eq. (1) and (2)
y1 - 2x21 - 3x1 + 9 - y1 + x21 + 3x1 = 0
- x21 + 9 = 0
x21 = 9
x1 = ± 3
when x1 = 3 y1 = 9 + 9 = 18
x1 = -3 y1 = 9 - 9 = 0
(-3 , 0) and (3, 18)
required point (-3, 0)
Let the point is P(x1, y1) on the curve
So,
= 2x + 3tangent
= 2x1 + 3(y - y1) = (x - x1) (2x1 + 3)
y - y1 = 2xx1 + 3x - 2x21 - 3x1
x(2x1 + 3) - y + y1 - 2x21 - 3x1 = 0
it passes (0, 9) then
0 (2x1 + 3) + 9 + y1 - 2x21 - 3x1 = 0
(x1, y1) also lies on the curve
= y1 - 2x21 - 3x1 + 9 = 0 ..... (1)
so y1 = x21 + 3x1 ..... (2)
From Eq. (1) and (2)
y1 - 2x21 - 3x1 + 9 - y1 + x21 + 3x1 = 0
- x21 + 9 = 0
x21 = 9
x1 = ± 3
when x1 = 3 y1 = 9 + 9 = 18
x1 = -3 y1 = 9 - 9 = 0
(-3 , 0) and (3, 18)
required point (-3, 0)
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