## Integer Type/Fill in The Blanks Type

Q.1₂₀₁₅ Suppose that the foci of the ellipse x²/9 + y²/5 = 1 are f1 and f2, where f1 > 0 and f2 < 0. Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and to (2f2, 0) respectively. Let T1 be the tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). The m is the slope of T1 and n is the slope of T2, then the value of (m⁻² + n²)...

Solution:
The equations of the parabolas P1 and P2 are y² = 8x and y² = -16x respectively
Tangent to y² = 8x passes through (-4, 0) ⇒ 0 = (-4m) + (2/m) ⇒ m⁻² = 2
And tangent to y² = -16x passes through (2, 0) ⇒ 0 = (2n) – (4/n) ⇒ n² = 2
⇒ m⁻² + n² = 4

Q.2₂₀₁₃ A vertical line passing through the point T(h, 0) intersect the ellipse x²/4 + y²/3 = 1 at the point P and Q. Let the tangents to the ellipse at P and Q meet at the point R. if Δ(h) = area of the triangle PQR, Δ₁ = maximum Δ(h) (where 1/2 ≤ h ≤ 1) and Δ₂ = minimum Δ(h) (where 1/2 ≤ h ≤ 1) then the value of (8Δ₁/√5) – 8Δ₂ is…

Solution:
For x = h, y = ± √(12 – 3h²)/2 ⇒ P = (h, √(12 – 3h²)/2) and Q = (h, – √(12 – 3h²)/2)
Let R(a, 0) be any point on the major axis of the ellipse
⇒ Chord of contact from R(a, 0) be ax/4 = 1 ⇒ x = 4/a ⇒ h = 4/a ⇒ a = 4/h
⇒ Area of triangle PQR is (PQ x RT)/2 = (√3(4 – h²)³/²)/2h
⇒ Δ₁ = maximum Δ(h) = (45√5)/8 {occurs at h = 1/2}
and Δ₂ = minimum Δ(h) = 9/2 {occurs at h = 1}
so the value of (8Δ₁/√5) – 8Δ₂ is (8/√5)((45√5)/8) – 8(9/2) = 45 – 36 = 9

Q.3₁₉₉₇ An ellipse has OB as a semi minor axis, F and F’ are its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is…

Solution:
Let the equation of the ellipse be x²/a² + y²/b² = 1
⇒ the coordinates of B, F , and F’ are (0, b), (ae, 0), and (-ae, 0) respectively (where e is the eccentricity of the ellipse)
since the angle FBF’ is a right angle ⇒ slope (BF).slope (BF’) = -1
⇒ (-b/ae)(b/ae) = – 1 ⇒ b² = (ae)²
⇒ (ae)² = a²(1 – e²) ⇒ e = (1/√2)

Q.4₁₉₉₆ An ellipse has eccentricity 1/2 and one focus at the point P(1/2, 1). It’s one directrix is the common tangent, nearer to the point P to the circle x² + y² = 1 and the hyperbola x² – y² = 1. The equation of the ellipse in the standard form is…

Solution:
The common tangent, nearer to the point P to the circle x² + y² = 1 and the hyperbola x² – y² = 1 is x = 1
so the directrix, focus and eccentricity of the ellipse are x = 1, (1/2, 1), and 1/2 respectively
so the equation of the ellipse is (x – 1/2)² + (y – 1)² = (1/2)²(x – 1)²
Now the equation of the ellipse in the standard form is (x – 1/3)²/(1/9) + (y – 1)²/(1/12) = 1

Q.5₁₉₉₄ Let P be a variable point on the ellipse x²/a² + y²/b² = 1 with foci F1 and F2. If A is the area of the triangle PF1F2, then the maximum value of A is…

Solution:
Let the coordinates of P, F1, and F2 are (aCosθ, bSinθ), (ae, 0), and (-ae, 0) respectively
⇒ area(A)of the triangle PF1F2 is = (ae)(bSinθ)
⇒ area is maximum at θ = π/2
⇒ A₍ₘₐₓ₎ = b√(a² – b²)