Example 1 :
Find the equations of those tangents to the circle
x^{2} + y^{2} = 52
which are parallel to the straight line 2x + 3y = 6.
Solution :
If two lines are parallel, then slopes of those lines will be equal.
Here,
slope of tangent = slope of the given line
Slope of the given line 2x + 3y = 6
m = -coefficient of x/coefficient of y
m = -2/3 ----(1)
Slope of the tangent line drawn to the circle
2x + 2y(dy/dx) = 0
dy/dx = -x/y ----(2)
(1) = (2)
-2/3 = -x/y
2y = 3x
y = 3x/2
Substituting the value of y in the equation of curve.
x^{2} + y^{2} = 52
x^{2 }+ (3x/2)^{2} = 52
x^{2 }+ (9x^{2}/4) = 52
13x^{2}/4 = 52
x^{2} = 16
x = ± 4
If x = 4, then y = 6 |
If x = -4, then y = -6 |
So, the point of contact are (4, 6) (-4, -6).
Equation of the tangent passing through the point (4, 6) and m = -2/3.
y - y_{1} = m(x - x_{1})
y - 6 = (-2/3)(x - 4)
3(y - 6) = -2(x - 4)
3y - 18 = -2x + 8
2x + 3y - 8 - 18 = 0
2x + 3y - 26 = 0
Equation of the tangent passing through the point (-4, -6) and m = -2/3.
y + 6 = (-2/3)(x + 4)
3(y + 6) = -2(x + 4)
3y + 18 = -2x - 8
2x + 3y + 26 = 0
Example 1 :
Find the equations of normal to
y = x^{3 }- 3x
that is parallel to 2x + 18y - 9 = 0.
Solution :
Slope of tangent drawn to the circle.
dy/dx = 3x^{2 }- 3(1)
= 3x^{2 }- 3
Slope of the normal line to the curve
= -1/(3x^{2 }- 3) ----(2)
Slope of the line parallel to the normal line to the curve
m = -coefficient of x/coefficient of y
m = -2/18
m = -1/9 ----(1)
(1) = (2)
-1/9 = -1/(3x^{2 }- 3)
3x^{2 }- 3 = 9
x^{2} = 12/3
x^{2 }= 4
x = √4
x = ±2
x = 2 y = 2^{3 }- 3(2) y = 2 |
x = -2 y = (-2)^{3 }- 3(-2) y = -2 |
points of contact are (2, 2) (-2, -2).
Equation of the normal passing through the point
(2, 2) and m = -1/9
y - 2 = (-1/9)(x - 2)
9(y - 2) = -1(x - 2)
9y - 18 = -x + 2
x + 9y - 20 = 0
Equation of the normal passing through the point
(-2, -2) and m = -1/9
y + 2 = (-1/9)(x + 2)
9(y + 2) = -1(x + 2)
9y + 18 = -x - 2
x + 9y + 20 = 0
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