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| PAGE: 143 | SET: Exercises | PROBLEM: 86 |
(a)
The equation of parabola is 
and the point is 
.
Slope of the tangent is derivative of the curve.
.
Apply derivative on each side with respect to 
.
Slope of the tangent is 
.
Point-slope form of line equation : 
.
Substitute 
and 
in the above formula.
This is a pair of tangent lines.
These tangent lines intersect the parabola, and the intersecting points can be determined by solving them.
Substitute 
in the curve .
and 
.
Substitute values in .
If 
, then 
.
If , then 
.
Therefore, the points at tangent lines intersect parabola are 
and 
.
Tangent line passing through 
:
.
Find the slope at .
.
Point-slope form of line equation : .
Substitute 
and 
in the above formula.
Tangent line passing through 
:
.
Find the slope at .
.
Point-slope form of line equation : .
Substitute 
and 
in the above formula.
(b)
The equation of parabola is and the point is 
.
Slope of the tangent to parabola is .
At 
, 
.
Assume that at 
is the tangent point.
Slope of the tangent line at 
is 
.
Slope of the line passing through two points 
and 
is defined as 
.
Here 
and 
Discriminant of quadratic equation 
is 
.
Here 
and 
.
Since the discriminant is negative there is no real values of 
.
Therefore there is no tangent line at 
.
Graph:
Graph the curve with the point .
.gif)
Tangents are 
and 
.
Graph is
.
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