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secondary 4 | A Maths
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Date Posted: 7 years ago
Views: 457
You have the gradient function of the curve. Substitute x value of point P gives you the gradient of the tangent at P. The normal is the inverse of of the tangent at P.
This normal is a straight line that cuts through the tangent of point P perpendicularly. You have the gradient of this line (inverse of the gradient of tangent at P) and the coordinates of point P itself. Substitute into y-mx+c to get the whole equation.
By substituting y=0, you can get the coordinates of Q.
You have the dy/dx equation. From there, you can integrate to get the equation of the curve (y and x). Substitute point P to get the value of C. You now have the equation of the curve
This normal is a straight line that cuts through the tangent of point P perpendicularly. You have the gradient of this line (inverse of the gradient of tangent at P) and the coordinates of point P itself. Substitute into y-mx+c to get the whole equation.
By substituting y=0, you can get the coordinates of Q.
You have the dy/dx equation. From there, you can integrate to get the equation of the curve (y and x). Substitute point P to get the value of C. You now have the equation of the curve
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Key concepts in this question are:
a) Normal of a point
b) Integrating via substitution method
If you have any questions, please do contact me! :)
a) Normal of a point
b) Integrating via substitution method
If you have any questions, please do contact me! :)
Date Posted:
7 years ago
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