parabola conic section

Question

from the point (4,6) a pair of tangent lines are drawn to the parabola y2 = 8x the area of triangle formed by these pair of tangent lines and the chord of contact of the point (4,6) is

Answer 1

\(A=\frac{(y_1^2-4ax_1)^{3/2}}{2a}\)

\(x_1=4;y_1=6;a=2\)

Answer 2

Equation of Tangent to parabola y² = 4ax is  y = mx + a/m

Then Equation of Tangent to parabola y² = 8x is  y = mx + 2/m 

This tangent is passing through (4, 6)

then  6 = 4m + 2/m

4m2 -6m + 2 = 0

4m2 -4m -2m +2 = 0

4m( m-1 ) -2 ( m-1 ) = 0

(2m -1) (m -1 ) then m = 1 and 1/2

Equation of tangents with above slopes

y = x + 2 --- Eqn 1

y = 1/2 x + 2(2)

x - 2y+ 8 = 0 -- Eqn2

Intersection of tangent y = x + 2 with  y² = 8x  is A( 2, 4)

Intersection of tangent x - 2y+ 8 = 0 with  y² = 8x  is B( 8, 8)

P( 4, 6), A( 2. 4), B(8,8)
translating axes to (-2, -4)

P( 2, 2), A( 0. 0), B(6,4)

Form a Determanant
|   2  2  1  |
|   0  0  1  |
|   6  4  1  |

Area of triangle PAB = Modulus [-(1/2) { - 8 + 12} ]= 2