A-Level Definite integrals

Method: the tangent at P and the chord PQ are the SAME straight line, so their gradients must be equal. Get one gradient by parametric differentiation and the other from the two points, then solve for q. Step 1 - gradient of the tangent at P, using dy/dx = (dy/dp) / (dx/dp). With x = ap² and y = a/p = ap⁻¹: dx/dp = 2ap dy/dp = −ap⁻² = −a/p² dy/dx = (−a/p²)/(2ap) = −1/(2p³) So the tangent at P has gradient −1/(2p³). Step 2 - gradient of the chord joining P(ap², a/p) and Q(aq², a/q), using (change in y)/(change in x). gradient = (a/q − a/p)/(aq² − ap²) Step 3 - simplify the numerator and denominator separately before dividing. Numerator: a/q − a/p = a(p − q)/(pq) Denominator: aq² − ap² = a(q² − p²) = a(q − p)(q + p) So the chord gradient is [a(p − q)/(pq)] ÷ [a(q − p)(q + p)] = (p − q) / [pq(q − p)(q + p)] Since p − q = −(q − p), the factor (q − p) cancels and leaves chord gradient = −1/[pq(p + q)] Step 4 - the tangent IS the chord, so equate the two gradients. −1/(2p³) = −1/[pq(p + q)] Taking reciprocals (or cross-multiplying): 2p³ = pq(p + q) Step 5 - divide both sides by p, which is allowed since p is not 0 (p = 0 gives no point on the curve because a/p would be undefined). 2p² = q(p + q) = pq + q² Step 6 - collect on one side and factorise as a quadratic in q. q² + pq − 2p² = 0 Looking for two terms multiplying to −2p² and adding to p gives 2p and −p: (q + 2p)(q − p) = 0 so q = −2p or q = p. Step 7 - reject q = p, because that is the point P itself, and Q is stated to be a SECOND point of intersection. Hence q = −2p.

Step 1: Differentiate the equation of C and evaluate dy/dx at each of the two points on C with x = a. Because C is symmetric about the x-axis over this stretch, those points are (a, a) and (a, -a), and the gradients work out as -1/2 at (a, a) and +1/2 at (a, -a). Step 2: Tangent at (a, a) with gradient -1/2, using y - y₁ = m(x - x₁). y - a = -(1/2)(x - a) y - a = -x/2 + a/2 y = -x/2 + a/2 + a y = -x/2 + 3a/2 Step 3: Tangent at (a, -a) with gradient +1/2. y - (-a) = (1/2)(x - a) y + a = x/2 - a/2 y = x/2 - a/2 - a y = x/2 - 3a/2 Step 4: The acute angle between two lines of gradients m₁ and m₂ satisfies tanθ = |(m₂ - m₁)/(1 + m₁m₂)| Here m₁ = -1/2 and m₂ = 1/2. Numerator: m₂ - m₁ = 1/2 - (-1/2) = 1 Denominator: 1 + m₁m₂ = 1 + (-1/2)(1/2) = 1 - 1/4 = 3/4 tanθ = |1 ÷ 3/4| = 4/3 Step 5: Take the inverse tangent. θ = arctan(4/3) = 53.130...° So the acute angle is 53.1° (3 s.f.), which is 0.927 rad (3 s.f.). (Useful check: the tangents meet where -x/2 + 3a/2 = x/2 - 3a/2, i.e. x = 3a and y = 0, so they cross on the x-axis at (3a, 0), and the two gradients ±1/2 are mirror images in that axis, which is why the angle is exactly twice arctan(1/2).)

Step 1: find where the normal meets the ellipse. Write the normal as y = √3(2x − 1/2) and substitute into (x−1)² + 4y² = 4: (x−1)² + 4 × 3(2x − 1/2)² = 4 (x² − 2x + 1) + 12(4x² − 2x + 1/4) = 4 x² − 2x + 1 + 48x² − 24x + 3 = 4 49x² − 26x = 0 x(49x − 26) = 0, so x = 0 or x = 26/49. At x = 0: y = 2√3(0) − √3/2 = −√3/2. This point (0, −√3/2) is on the ellipse and on the y-axis, so all three boundaries of R meet there. At x = 26/49: y = 2√3(26/49) − √3/2 = √3(104 − 49)/98 = 55√3/98. So the normal cuts the ellipse again at P(26/49, 55√3/98) = (0.530612, 0.972069). Step 2: write the boundaries with x as a function of y, since the rotation is about the y-axis. Normal: y = 2√3x − √3/2 rearranges to x = (y + √3/2)/(2√3). Ellipse: (x−1)² = 4 − 4y², so x = 1 ± 2√(1 − y²). The arc bounding R is the left-hand branch, x = 1 − 2√(1 − y²). The ellipse meets the y-axis where x = 0, i.e. 1 + 4y² = 4, giving y = ±√3/2; the upper of these is (0, √3/2). So for −√3/2 ≤ y ≤ √3/2 the strip runs from x = 0 out to the normal, and for √3/2 ≤ y ≤ 55√3/98 it runs from the ellipse out to the normal. Step 3: set up the volume as a difference of two rotations. V = pi × INT from −√3/2 to 55√3/98 of [(y + √3/2)/(2√3)]² dy − pi × INT from √3/2 to 55√3/98 of [1 − 2√(1 − y²)]² dy. Step 4: evaluate the first term. Rotating the straight line about the y-axis sweeps a cone with apex at (0, −√3/2), base radius 26/49 and height 55√3/98 + √3/2 = √3(55 + 49)/98 = 52√3/49. So the first term = (1/3)pi(26/49)²(52√3/49) = 0.541940 (6 d.p.). Step 5: evaluate the second term with the calculator: pi × INT from √3/2 to 55√3/98 of [1 − 2√(1 − y²)]² dy = 0.024821 (6 d.p.). Step 6: subtract. V = 0.541940 − 0.024821 = 0.517119 = 0.517 cubic units (3 significant figures).