Q.276. Area of quadrilateral formed by common tangents of parabola and ellipse

Question 276.
Common tangents are drawn to the parabola y^2 = 4x and the ellipse 3x^2 + 8y^2 = 48 touching the parabola at A and B and the ellipse at C and D. Find the area of the quadrilateral.

Answer 276.
P (t^2, 2t) is any point on the parabola y^2 = 4x.
The equation of tangent at P to the parabola is
2ty = 2(x + t^2)
=> ty = x + t^2
=> x = ty - t^2

If this line is also a tangent to the ellipse, its solution with the equation of the ellipse must be a unique point. Solving with the equation of the given ellipse,
3 (ty - t^2)^2 + 8y^2 = 48
=> (3t^2 + 8) y^2 - 6t^3 y + 3t^4 - 48 = 0 ... (1)

As y is unique, discriminant of this quadratic should be zero
=> 36t^6 - 4 (3t^2 + 8) (3t^4 - 48) = 0
=> 3t^6 - (3t^2 + 8) (t^4 - 16) = 0
=> - 8t^4 + 48t^2 + 128 = 0
=> t^4 - 6t^2 - 16 = 0
=> (t^2 - 8) (t^2 + 2) = 0
=> t^2 = 8
=> t = 2√2
=> A = (8, 4√2) and B = (8, - 4√2)
                  ... [Note: B is symmetric with A about x-axis.]

Plugging the value of t in eqn. (1),
32y^2 - 96√2 y + 144 = 0
=> 4y^2 - 12√2y + 18 = 0
=> (2y - 3√2)^2 = 0
=> y = (3√2) / 2
Also, x = ty - t^2 = (2√2) * (3√2)/2 - 8 = - 2
=> C = (- 2, (3√2) / 2) and D = (- 2, (3√2) / 2)
                ... [Note: D is symmetric with C about x-axis.]

Sides AB and CD are parallel
=> CABD is a trapezium
=> its area
= (1/2) altitude x (sum of lengths of parallel sides)
= (1/2) * difference of x-coordinates x (3√2 + 8√2)
= (1/2) * (8 - (- 2)) * 11√2
= 55√2 sq. units.

Link to YA!