a. The equation of the tangent at (1,3) is y = -x/3 + 10/3
b. The equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
a. How to find the equation of the tangent at (1, 3)?
Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.
So, x² + y² = 10
d(x² + y²)/dx = d10/dx
dx²/dx + dy²/dx = 0
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y
At (1,3), dy/dx = -1/3
Using the equation of a straight line in slope-point form, we have
m = (y - y₁)/(x - x₁) where
- m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
- (x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
-1/3 = (y - 3)/(x - 1)
-(x - 1) = 3(y - 3)
-x + 1 = 3y - 9
3y = -x + 1 + 9
3y = -x + 10
3y + x = 10
y = -x/3 + 10/3
So, the equation of the tangent at (1,3) is y = -x/3 + 10/3
b. The equation of the normal at the point (1, 3)
Since the tangent and normal line are perpendicular at the point, for two perpendicular line,
mm' = -1 where
- m = gradient of tangent = -1/3 and
- m' = gradient of normal
So, m' = -1/m
= -1/(-1/3)
= 3
Using the equation of a straight line in slope-point form, we have
m' = (y - y₁)/(x - x₁) where
- m' = gradient of normal at (1, 3) and (
- x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
3 = (y - 3)/(x - 1)
3(x - 1) = (y - 3)
3x - 3 = y - 3
y = 3x - 3 + 3
y = 3x + 0
y = 3x
So, the equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
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