3.
Find the gradient of the tangent by differentiating implicitly Differentiating both sides with respect to x y We will use the general equation of a line At the point P x
4.
So the GRADIENT OF THE TANGENT AT POINT P IS The GRADIENT OF THE NORMAL IS THEREFORE Because the product of the gradients of TANGENT and NORMAL is -1
5.
This is the required EQUATION OF THE NORMAL QED We will use the general equation of a line
6.
The normal meets the x axis at Q when y=0 find the x coordinate on the normal Q is the point
7.
R is the midpoint of PQ FIND THE MIDPOINT X Coordinate of R Y Coordinate of R (THE MIDPOINT) R is
8.
To find the LOCUS OF R as p varies we ELIMINATE p and SUBSTITUTE This is the equation of the LOCUS of R as p varies.
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