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 = at2 , 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 y2 = 8x . (Hint: use parametric form)
Solution :
x = at2, y = 2at
y2 = 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 :
yy1 = 8[(x + x1)/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 12x2 − 9y2 = 108 at θ = π/3 . (Hint : use parametric form)
Solution :
x = a sec θ , y = b tan θ
(12x2/108) − (9y2/108) = 108/108
(12x2/108) − (9y2/108) = 108/108
(x2/9) − (y2/12) = 1
a2 = 9, b2 = 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(xx1) − 9(yy1) = 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 ‘ t1 ’ and ‘ t2 ’on the parabola y2 = 4ax is a t1 t2 , a(t1 + t2)
If the normal at the point ‘ t1 ’ on the parabola y2 = 4ax meets the parabola again at the point ‘ t2 ’, then prove that t2 = -(t1 + 2/t1)
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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