Question

Tangents drawn to the parabola y^2=8x at the points p(t1) and q(t2) intersect at a point r such that t1 and t2 are the roots of the equation t^2+at+2=0, |a|>2root2. find locus of circumcenter of triangle ptq is

Answer 1

Solution:

Given equation: y² = 8x         ... (1)

The above equation compare the general equation of parabola y² = 4ax

So, a = 2.

The parametric coordinates of any point on (1) are (at², 2at) or (2t², 4t).

Coordinates of the point P are P(2t₁². 4t₁) and Q are Q(2t₂². 4t₂).

Differentiate equation (1),

2y dy/dx= 8

dy/dx= 4/y

Slope of the tangent at point P = 4/4t₁  = 1/t₁

Slope of the tangent at point Q = 4/4t₂  = 1/t₂

Now, the equation of RP is,

(y - 4t₁)/(x - 2t₁²) =  1/t₁

t₁y - 4t₁² = x - 2t₁²

t₁y - x = 2t₁²      ... (2)

Now, the equation of RQ is,

t₂y - x = 2t₂²     ... (3)

Solve equation (2) and (3),

x = 2t₁t₂; y = 2(t₁ + t₂)

Now, t² + at + 2 = 0

Since, t₁ and t₂ are the roots of the above equation,

t₁ + t₂ = -a

t₁t₂ = 2

so, x = 4; y = -2a

Coordinates of the point R is (4, -2a)

Hence, the circumcentre of ΔPQR is ((4 + 2t₁²+2t₂² )/2, (-2a+4t₁ + 4t₂)/2) ≡ (a²-2, -3a)

Answer 2

Given equation: y² = 8x         ... (1)

The above equation compare the general equation of parabola y² = 4a

So, a = 2.

The parametric coordinates of any point on (1) are (at², 2at) or (2t², 4t).

Coordinates of the point P are P(2t₁². 4t₁) and Q are Q(2t₂². 4t₂).

Differentiate equation (1),

2y dy/dx= 8

dy/dx= 4/y

Slope of the tangent at point P = 4/4t₁  = 1/t₁

Slope of the tangent at point Q = 4/4t₂  = 1/t₂

Now, the equation of RP is,

(y - 4t₁)/(x - 2t₁²) =  1/t₁

t₁y - 4t₁² = x - 2t₁²

t₁y - x = 2t₁²      ... (2)

Now, the equation of RQ is,

t₂y - x = 2t₂²     ... (3)

Solve equation (2) and (3),

x = 2t₁t₂; y = 2(t₁ + t₂)

Now, t² + at + 2 = 0

Since, t₁ and t₂ are the roots of the above equation,

t₁ + t₂ = -a

t₁t₂ = 2

so, x = 4; y = -2a

Coordinates of the point R is (4, -2a)

Hence, the circumcentre of ΔPQR is ((4 + 2t₁²+2t₂² )/2, (-2a+4t₁ + 4t₂)/2) ≡ (a²-2, -3a)