Question

Let PQ be a focal chord of the parabola y² = 4x such that it subtends an angle of `π/2` at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: `x^2/a^2 + y^2/b^2` = 1, a² > b². If e is the eccentricity of the ellipse E, then the value of `1/e^2` is equal to ______.

Options

• `1 + sqrt(2)`

• `3 + 2sqrt(2)`

• `1 + 2sqrt(3)`

• `4 + 5sqrt(3)`

Answer 1

Let PQ be a focal chord of the parabola y² = 4x such that it subtends an angle of `π/2` at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: `x^2/a^2 + y^2/b^2` = 1, a² > b². If e is the eccentricity of the ellipse E, then the value of `1/e^2` is equal to `underlinebb(3 + 2sqrt(2))`.

Explanation:

Given: Focal chord of y² = 4x is PQ and it subtends an angle of `π/2` at point (3, 0)

Let parametric coordinates of point P be (t², 2t) then parametric coordinates of point Q be `(1/t^2,(-2)/t)`

∵ AP ⊥ AQ

∴ (m_AP)(m_AQ) = –1

⇒ `((2t)/(t^2 - 3))(((-2)/t)/(1/t^2 - 3))` = –1

⇒ `(-4t^2)/((t^2 - 3)(1 - 3t^2))` = –1

⇒ 4t² = –3t⁴ + 10t² – 3

⇒ 3t⁴ – 6t² + 3 = 0

⇒ (t² – 1)² = 0

⇒ t = 1

∴ Coordinates of point P and Q are (1, 2) and (1, –2) respectively.

Since, line segment PQ is also a focal chord of ellipse `x^2/a^2 + y^2/b^2` = 1, a² > b².

∴ P and Q must be end points of latus rectum

⇒ `(2b^2)/a` = 4 and ae = 1

As we know, b² = a²(1 – e²)

⇒ b² = a² – a²e²

⇒ b² = a² – 1

⇒ `(a^2 - 1)/a` = 2

⇒ a² – 2a – 1 = 0

⇒ a = `1 + sqrt(2)`

⇒ e = `1/a = 1/(1 + sqrt(2))`

⇒ e² = `1/(3 + 2sqrt(2))`

⇒ `1/e^2 = 3 + 2sqrt(2)`