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The equation of the tangent to the curve (1 + x2)y = 2 - x where it crosses x-axis is
- a)x + 5y = 2
- b)x - 5y = 2
- c)5x - y = 2
- d)5x + y - 2 = 0
Correct answer is option 'A'. Can you explain this answer?
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The equation of the tangent to the curve (1 + x2)y = 2 - x where it cr...
Given equation: (1+x^2)y = 2-x
To find the tangent to the curve at x-axis, we need to find the derivative of the curve at that point.
Differentiating both sides with respect to x, we get:
(1+x^2)dy/dx + 2xy = -1
Now at x-axis, y=0, so we get:
2xy = -1
x = -1/2y
Substituting this value of x in the equation of the curve, we get:
(1 + (−1/2y)^2)y = 2 − (−1/2y)
(1 + 1/4y^2)y = 2 + 1/2y
y^3 + 4y - 8 = 0
Differentiating this with respect to y, we get:
3y^2 + 4 = 0
y^2 = -4/3
Since y cannot be imaginary, we reject this value and conclude that there is no tangent to the curve at x-axis.
Therefore, none of the given options are correct.
Note: There might be a typo/mistake in the original question.
To find the tangent to the curve at x-axis, we need to find the derivative of the curve at that point.
Differentiating both sides with respect to x, we get:
(1+x^2)dy/dx + 2xy = -1
Now at x-axis, y=0, so we get:
2xy = -1
x = -1/2y
Substituting this value of x in the equation of the curve, we get:
(1 + (−1/2y)^2)y = 2 − (−1/2y)
(1 + 1/4y^2)y = 2 + 1/2y
y^3 + 4y - 8 = 0
Differentiating this with respect to y, we get:
3y^2 + 4 = 0
y^2 = -4/3
Since y cannot be imaginary, we reject this value and conclude that there is no tangent to the curve at x-axis.
Therefore, none of the given options are correct.
Note: There might be a typo/mistake in the original question.
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The equation of the tangent to the curve (1 + x2)y = 2 - x where it crosses x-axis isa)x + 5y = 2b)x - 5y = 2c)5x - y = 2d)5x + y - 2 = 0Correct answer is option 'A'. Can you explain this answer?
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