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The gradient of the tangent line at the point (a cos a, a sin a) to the circle x2 + y2 = a2, is
- a)tan(p–a)
- b)tan a
- c)cot a
- d)– cot a
Correct answer is option 'D'. Can you explain this answer?
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To find the gradient of the tangent line at a point on a circle, we need to find the derivative of the circle equation with respect to x and y.
Given the circle equation x^2 + y^2 = a^2, we can differentiate both sides with respect to x to get:
2x + 2y * dy/dx = 0
Simplifying, we have:
dy/dx = -x/y
At the point (a cos a, a sin a), x = a cos a and y = a sin a. Substituting these values into the derivative equation, we have:
dy/dx = -(a cos a) / (a sin a)
Simplifying further, we get:
dy/dx = -cot a
Therefore, the gradient of the tangent line at the point (a cos a, a sin a) to the circle x^2 + y^2 = a^2 is -cot a.
Given the circle equation x^2 + y^2 = a^2, we can differentiate both sides with respect to x to get:
2x + 2y * dy/dx = 0
Simplifying, we have:
dy/dx = -x/y
At the point (a cos a, a sin a), x = a cos a and y = a sin a. Substituting these values into the derivative equation, we have:
dy/dx = -(a cos a) / (a sin a)
Simplifying further, we get:
dy/dx = -cot a
Therefore, the gradient of the tangent line at the point (a cos a, a sin a) to the circle x^2 + y^2 = a^2 is -cot a.
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The gradient of the tangent line at the point (a cos a, a sin a) to the circle x2+ y2= a2, isa)tan(p–a)b)tan ac)cot ad)–cot aCorrect answer is option 'D'. Can you explain this answer?
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