Problem 1 :
Find the equation of the two tangent can be drawn from (5, 2) to the ellipse 2x^{2} + 7y^{2} = 14
Solution :
2x^{2} + 7y^{2} = 14
Equation of tangent line from the point (x, y) will be
y = mx ±√a^{2}m^{2}+b^{2}
Equation of tangent :
While applying the values of m, we get
Problem 2 :
Find the equations of tangents to the hyperbola
which are parallel to 10x - 3y + 9 = 0
Solution :
a^{2} = 16, b^{2} = 64
From the given line, we find the value of m.
3y = 10x + 9
y = (10/3)x + 3
m = 10/3
Problem 3 :
Show that the line x - y + 4 = 0 is a tangent to the ellipse
x^{2} + 3y^{2} = 12
Also find the coordinate of the point of contact.
Solution :
If the line y = m x + b is a tangent to the given ellipse it has to satisfy the following condition.
c^{2} = a^{2}m^{2} + b^{2}
x - y + 4 = 0
y = x + 4
m = 1 and c = 4
4^{2} = 12(1)^{2} + 4
16 = 12 + 4
16 = 16
So, the given line is tangent to the given ellipse.
Problem 4 :
Find the equation of the tangent to the parabola y^{2} = 16x perpendicular to 2x + 2y + 3 = 0
Solution :
2x + 2y + 3 = 0
2y = -2x - 3
y = -x - (3/2)
m = -1
Slope of the tangent line drawn to the curve = 1
y^{2} = 16x can be compared with y^{2} = 4ax
4a = 16 and a = 4
Equation of tangent line :
y = mx +(a/m)
y = 1x + (4/1)
y = x + 4
Problem 5 :
Find the equation of the tangent at t = 2 to the parabola y^{2} = 8x
Solution :
y^{2} = 8x
Point of the tangent at t = 2 for a parabola will be (at^{2}, 2at)
y^{2} = 8x
y^{2} = 4ax
4a = 8 and a = 2
Finding at^{2} a = 2 and t = 2 = 2(2)^{2} = 8 |
Finding 2at a = 2 and t = 2 = 2(2)(2) = 8 |
Equation of the tangent to the parabola :
y = mx + a/m ---(1)
Applying (8, 8) to find m.
8 = m(8) + 2/m
8 - 8m = 2/m
8m - 8m^{2} = 2
4m - 4m^{2} = 1
4m^{2} - 4m + 1 = 0
(2m - 1) (2m - 1) = 0
m = 1/2 and m = 1/2
Applying m = 1/2 in (1), we get
y = (1/2)x + 2/(1/2)
y = x/2 + 4
2y = x + 8
x - 2y + 8 = 0
Problem 6 :
Find the equations of tangent and normal to hyperbola 12x^{2} - 9y^{2} = 108 at θ = π/3
Solution :
Equation of tangent for the parabola :
12xx_{1} - 9yy_{1} = 108
At (6, 6)
12x(6) - 9y(6) = 108
72x - 54y = 108
Dividing by 6, we get
12x - 9y = 18
Dividing by 3, we get
4x - 3y = 6
Equation of normal :
3x + 4y = k
At (6, 6)
3(6) + 4(6) = k
k = 18 + 24
k = 42
3x + 4y = 42
So, equation of tangent is 4x - 3y = 6 and normal is 3x + 4y = 42.
Problem 7 :
Prove that the point of intersection of the tangents at t_{1} and t_{2} on the parabola y^{2} = 4ax is (at_{1} t_{2}, a(t_{1} + t_{2}))
Solution :
Equation of tangent at the point t_{1} and t_{2} :
Equation of tangent for the parabola
yt = x + at^{2}
Equation of tangent at the point t_{1} and t_{2} are yt_{1} = x + at_{1}^{2 }and yt_{2} = x + at_{2}^{2}
yt_{1} = x + at_{1}^{2 }---------(1)
yt_{2} = x + at_{2}^{2 }---------(2)
(1) - (2)
yt_{1} - yt_{2 }= x + at_{1}^{2 }- (x + at_{2}^{2})
y(t_{1} - t_{2}) = at_{1}^{2 }- at_{2}^{2}
y(t_{1} - t_{2}) = a(t_{1}^{2 }- t_{2}^{2})
y(t_{1} - t_{2}) = a(t_{1 }+ t_{2}) (t_{1 }- t_{2})
y = a(t_{1 }+ t_{2})
Applying the value of y in (1)
a(t_{1 }+ t_{2})t_{1} = x + at_{1}^{2 }
a t_{2}t_{1} = x
x = a t_{1}t_{2}
Point of intersection is (a t_{1}t_{2, }a(t_{1 }+ t_{2})).
Problem 8 :
If the normal at the point t_{1} on the parabola y^{2} = 4ax meets the parabola again at the point t_{2} , then prove that
t_{2} = -(t_{1} + 2/t_{1})
Solution :
Equation of normal to the parabola is
y + xt = 2at + at^{3}
It passes through t_{1}.
y + x t_{1} = 2a t_{1} + a t_{1}^{3 }-------(1)
Any point on the parabola will be in the form (at^{2}, 2at)
It passes through t_{2}.
(at^{2}, 2at) ==> (at_{2}^{2}, 2at_{2})
2at_{2} + at_{2}^{2} t_{1} = 2a t_{1} + a t_{1}^{3 }------(2)
2at_{2} - 2a t_{1} = a t_{1}^{3}- at_{2}^{2} t_{1}
2a(t_{2} - t_{1}) = at_{1} (t_{1}^{2}- t_{2}^{2})
2a(t_{2} - t_{1}) = -at_{1} (t_{2}^{2}- t_{1}^{2})
2a(t_{2} - t_{1}) = -at_{1} (t_{1}- t_{2})(t_{1}+ t_{2})
2a= -a t_{1} (t_{1}+ t_{2})
2 = - t_{1} (t_{1}+ t_{2})
(t_{1}+ t_{2}) = -2/ t_{1}
t_{2 }= -2/t_{1} - t_{1}
t_{2 }= -(2/t_{1} + t_{1})
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM