Find the Equation of the Tangent to the Parabola in Parametric Form :
Here we are going to see some practice questions to find the equation of the tangent to the parabola in parametric form.
Name of the conic Circle Parabola Ellipse Hyperbola |
Parametric equations x = a cos θ , y = a sin θ x = at^{2} , y = 2at x = a cos θ , y = b sin θ x = a sec θ , y = b tan θ |
Question 1 :
Find the equation of the tangent at t = 2 to the parabola y^{2} = 8x . (Hint: use parametric form)
Solution :
x = at^{2}, y = 2at
y^{2} = 8x
4a = 8 ==> a = 2
x = 2(2)^{2} ==> 8
y = 2(2)(2) ==> 8
The point on the tangent line is (8, 8).
Equation of tangent to the parabola :
yy_{1} = 8[(x + x_{1})/2]
y(8) = 4(x + 8)
8y = 4x + 32
4x - 8y + 32 = 0
Divide the equation by 4, we get
x - 2y + 8 = 0
Hence the required equation of the tangent line is x - 2y + 8 = 0.
Question 2 :
Find the equations of the tangent and normal to hyperbola 12x^{2} − 9y^{2} = 108 at θ = π/3 . (Hint : use parametric form)
Solution :
x = a sec θ , y = b tan θ
(12x^{2}/108) − (9y^{2}/108) = 108/108
(12x^{2}/108) − (9y^{2}/108) = 108/108
(x^{2}/9) − (y^{2}/12) = 1
a^{2} = 9, b^{2} = 12
a = 3 and b = 2√3
x = a sec θ x = 3 sec π/3 x = 3(2) = 6 |
y = b tan θ y = 2√3 tan π/3 y = 2√3(√3) = 6 |
Equation of tangent :
12(xx_{1}) − 9(yy_{1}) = 108
12x(6) − 9y (6) = 108
72x - 54y = 108
8x - 6y = 12
4x - 3y - 4 = 0
Equation of normal :
3x + 4y + k = 0
3(6) + 4(6) + k = 0
18 + 24 + k = 0
k = -42
3x + 4y - 42 = 0
Prove that the point of intersection of the tangents at ‘ t_{1} ’ and ‘ t_{2} ’on the parabola y^{2} = 4ax is a t_{1} t_{2} , a(t_{1} + t_{2})
If the normal at the point ‘ t1 ’ on the parabola y2 = 4ax meets the parabola again at the point ‘ t2 ’, then prove that t_{2} = -(t_{1} + 2/t_{1})
After having gone through the stuff given above, we hope that the students would have understood, "Find the Equation of the Tangent to the Parabola in Parametric Form".
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