FIND THE EQUATION OF THE TANGENT TO THE PARABOLA IN PARAMETRIC FORM

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

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)

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