To find equation of the tangent line, we should be aware of the parametric form
ShapesCircle |
Parametric equationsx = a cos θ, y = a sin θ |
Parabola |
x = at^{2}, y = 2at |
Ellipse |
x = a cos θ, y = a sin θ |
Hyperbola |
x = a sec θ, y = b tan θ |
In the given conic function, we have to do the following changes.
x^{2} = xx_{1}, y^{2} = yy_{1}
x = (x + x_{1})/2 and y = (y + y_{1})/2
Applying the point that we have derived from parametric equation in this equation, we will get equation of the tangent line.
ax + by = c
To find the normal, we have to interchange the coefficients of x and y. That is,
bx - ay = k
Applying x and y, we can solve for k.
Find the equations of the tangent and normal
Problem 1 :
To the parabola y^{2} = 8x at t = 1/2
Solution :
Given, equation of the parabola y^{2} = 8x.
Comparing this equation with y^{2} = 4ax
Point of the tangent at t = 1/2 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 = 1/2 = 2(1/2)^{2} = 2(1/4) = 1/2 |
Finding 2at a = 2 and t = 1/2 = 2(2)(1/2) = 4/2 = 2 |
Equation of the tangent to the parabola :
y = mx + a/m ---(1)
Applying (1/2, 2) to find m.
2 = m(1/2) + 2/m
2 = m/2 + 2/m
2 = (m^{2} + 4)/2m
4m = m^{2} + 4
m^{2} - 4m + 4 = 0
(m - 2) (m - 2) = 0
m = 2 and m = 2
Applying m = 2 in (1), we get
y = 2x + 2/2
y = 2x + 1
2x - y + 1 = 0
Equation of normal :
2y = x + k
At(1, 1)
2(1) = 1 + k
2 = 1 + k
k = 1
2y = x + 1
x - 2y + 1 = 0
So, equation of tangent is 2x - y + 1 = 0 and normal is x - 2y + 1 = 0.
Problem 2 :
To the ellipse x^{2} + 4y^{2} = 32 at θ = π/4
Solution :
x^{2} + 4y^{2} = 32
Dividing 32 on each sides.
a^{2} = 32 a = √32 a = √(16 × 2) a = 4√2 x = a cos θ x = 4√2 cos (π/4) = 4√2 × (1/√2) x = 4 |
b^{2} = 8 b = √8 b = √(4 × 2) b = 2√2 y = b sin θ y = 2√2 sin (π/4) = 2√2 × (1/√2) y = 2 |
Equation of the tangent :
y - y_{1} = m(x - x_{1})
y - 2 = -1/2(x - 4)
y - 2 = -x/2 + 4/2
y = -x/2 + 4/2 + 2
y = -x/2 + 4/2 + 4/2
y = (-x + 8)/2
2y = -x + 8
2y + x = 8
x + 2y - 8 = 0
Equation of normal :
2x + y = k
At (4, 2)
2(4) + 2 = k
8 + 2 = k
10 = k
2x + y - 10 = 0
So, equation of tangent is x + 2y - 8 = 0 and normal is 2x - y - 10 = 0.
Problem 3 :
Solution :
a^{2} = 9 a = 3 x = a sec θ x = 3 sec (π/6) = 3 (2/√3) x = 2√3 |
b^{2} = 12 b= 2√3 y = b tan θ y = 2√3 tan (π/6) = 2√3 (1/√3) y = 2 |
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM