This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you'll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.
(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4
2x^2 + 5y^2 = 1.75
Now you have an ellipse in the z=1.5 plane as your trace.
(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
5y^2 + z^2 = 2
Now you have an ellipse in the x=1 plane as your trace.
(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
2x^2 + 5y^2 = 4
Now you have an ellipse in the z=0 plane as your trace.
(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
2x^2 + z^2 = 4
Now you have an ellipse in the y=0 plane as your trace.
Answer:
x=4, y=9.6
Step-by-step explanation:
Using Theorem of Intersecting Secant and Tangent


Next, we apply Theorem of Intersecting Chords
SV X VR=PV X VQ
5 X y = x X 12
Recall: x=4
5y=4 X 12
5y=48
y=48/5=9.6
Therefore: x=4, y=9.6
Answer:
The gradient of
is
.
Step-by-step explanation:
Geometrically speaking, the gradient of a curve is the slope of the tangent line at a given point of the curve. From perspective of Differential Calculus, the gradient is the first derivative of the curve. Let
, then the gradient of the curve is:
(1)
The gradient of
is
.
So We Have:
11i(5 + 8i)
We Need To Distribute.
We Get:
55 + 88i²