Mixed exercise 3. x y. cosh t sinh t 1 Substituting the values for cosh t and sinht in the equation for the hyperbola H. = θ =

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1 Mixed exercise x x a Parametric equations: cosθ and sinθ 9 cos θ + sin θ Substituting the values for cos θ and sinθ in the equation for ellipse E gives the Cartesian equation: + 9 b Comparing with the equation in its standard form, So a sketch of E is: +, a and b 9 a b 9cos c Use the chain rule to find the gradient: d θ θ sinθ dθ The gradient of the normal is sin θ 9cosθ sinθ Use y y0 m( x x0) to get the equation: y 9sin θ ( x cos θ) 9cosθ 9ycosθ 8sinθcosθ sin θ( x cos θ) So the equation of the normal is xsinθ 9ycosθ 6sinθcosθ a Parametric equations: x ± cosht and y sinht cosh t sinh t Substituting the values for cosh t and sinht in the equation for the hyperbola H gives the Cartesian equation: Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2 b Comparing with the equation in its standard form,, a and b a b So the asymptotes are the lines y ± x and the hyperbola cuts the x-axis at ± A sketch is the following: cosh c Use the chain rule to find the gradient: d t t sinht dt cosht Use y y0 m( x x0) to get the equation: y sinh t ( x cosh t) sinht ysinht + 0 xcosht b b a Asymptotes arey ± x so m b am a a Substituting in the equation for the hyperbola: a a m Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

3 b Let Q be the point on the asymptote. tanθ m θ y tan x θ y tan x xy Using tanθ m () θ ( ) tan y x But P lies on the hyperbola so from part a, x m y a m So m y x a Using () and () So () x ( x a ) ( x y ) ( x y ) y x a a Gradient of chord PQ c c p q cp cq c( q p) pqc( p q) pq c c c b P cp, ; Q cq, ; R cr, p q r Gradient of PQ pq Gradient of PR pr If QPR ˆ 90 p qr () pq pr To find gradient of tangent at P, let q p for chord PQ Gradient of tangent at P is p Gradient of chord RQ is qr qr p p qr But from () p qr So gradient of tangent at P gradient of QR Therefore the tangent at P is perpendicular to chord QR. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

4 a 6 a ct y ct, x ct c t c Equation of tangent is: y ( x ct ) t t yt ct x + ct or t y + x ct b Let S cs, c s and T ct, c be two points on the hyperbola 6( ) t xy c So tangent at S is s y + x cs Using the equations for the tangent at T and S, intersection of tangents is: ( t s ) y ct ( s) c y t + s ct cts So x ct t + s t + s 8ts 8 When c the point of intersection is, t + s t + s The tangents intersect at (, ): 8ts x : ( t + s) 8ts t + s 8 y : ( t + s) 8 t + s So ts t s Substituting for t: s 8 s 8s 0 s Factorising: (s + )( s ) 0 So s or s ; t or t The points S and T are, and, 9x + y + 9 a, b 6 b a ( e ) 9 ( e ) e e a The foci are at ±,0, which is ( ±, 0), so A and B are the foci. e From the properties of an ellipse, PS + PS a 0 So PA+ PB 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

5 6 b Normal at P is: xsint ycost 6costsint 6 X is when y 0, i.e. x cost PB (cost ) + (sin t) cos t 0cost sin t 6cos t 0cost + (cost ) PB cos t and PA 0 PB + cost 6 6 AX + cos t, BX cost If PX bisects angle APB, then angle θ angle φ 6 sin(80 α) AX sin α(+ cos t) Consider sine rule on PAX: sinφ sinα AP + cost 6 sinαbx sin α( cos t) Consider sine rule on PBX: sinθ sinα PB cost So sinθ sinφ and since both angles are acute, the normal bisects APB. 7 a ct y ct, x ct c t c Equation of tangent is: y ( x ct ) t t yt ct x + ct or t y + x ct b Gradient of tangent is, so gradient of OP is t t Equation of OP is y t x Equation of tangent is t y ct x Solving t x ct x ct ct x, y + t + t c t + c t c t (+ t x + y ( + t ) ( +t ) ( x + y ) 6 6c t ( + t ) ( ) c xy + c t xy ( ) + t ) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

6 ap 8 a Gradient OP, gradient of ap p Since OP and OQ are perpendicular, pq pq OQ q b Normal at Q is y + xq aq + aq c Normal at P is y + xp ap + ap Solving the equations for the tangents simultaneously: x( q p) a( q p ) + a( q p) x( q p) a( q p) x a q + p + qp + ( q + qp + p ) + a( q p) ( ) y ap + ap But if pq then R is d Express Then p + q as ( p+ q) pq apq ap aqp ap y apq( q+ p) ( ap + aq a, a( p+ q)) X a p+ q pq a p+ q + Y Y a( p + q) p + q a (( ) ) (( ) 6) Y X a + 6 6a Y X 6a Y 6aX 96a 6a 9 y mx + c and + a b b x + a ( mx + c) a b b x + a m x + a mxc + a c a b x ( b + a m ) + a mcx + a ( c b ) 0 For a tangent the discriminant 0: a m c ( b + a m ) a ( c b ) a m c b c b + a m c a m b a m b + b b c c a m + b So the lines c ± a m + b y mx ± a m + b are tangents for all m. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 0 Chord PQ has gradient c c p q c( q p) cp cq pqc( p q) pq If gradient, then pq Tangent at P is Tangent at Q is p y+ x cp q y + x cq c Intersection: ( p q ) y c( p q) y p + q cp cpq x cp p + q p + q cpq c So the point of intersection R is, p + q p + q c c But pq, so R is x, y p + q p + q, i.e. y x The locus of R is the line y x cos a Use the chain rule to find the gradient: dθ θ cosθ 6sinθ sinθ dθ cos θ Line l has equation: y sin θ ( x 6cos θ) sinθ ysinθ sin θ xcos θ+ cos θ θ+ y θ θ+ θ xcos sin cos sin xcos θ+ ysinθ 6 b y 0: xcosθ x cosθ x 0: ysinθ y sinθ 6 So A has coordinates,0 cosθ and B has coordinates 0, sinθ Midpoint of AB has coordinates, cosθ sinθ, so x and cosθ Using cos θ + sin θ : + x y 9 The locus of the midpoint of AB is + y sinθ Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 cos a Use the chain rule to find the gradient: d θ θ sinθ dθ cosθ An equation for the tangent line l is : y sin θ ( x cos θ) sinθ ysinθ 6sin θ xcosθ + 6cos θ + ysinθ xcos 6 b The point A has x-coordinate equal to 0, so its coordinates are 0, sinθ Line l is perpendicular to l so has gradient sin θ cosθ sinθ The equation of the line l is given by: y x sinθ cosθ ysinθcosθ cosθ xsin θ c Using a and b, b a ( e ) 69( e ) e 69 So the eccentricity of the ellipse is e l cuts the x-axis at ( ae, 0), which is (, 0). Substitute this into the equation of l : cosθ 6sin θ 6 cos θ + cosθ 6 0 cosθ 6( cos θ) ± The solutions of this equation arecosθ 8 This gives either cosθ, which can t be a cosine, or cosθ e 8sec a Use the chain rule to find the gradient: d θ θ secθtanθ sinθ dθ sinθ The gradient of the normal l is sinθ Use y y0 m( x x0) to get the equation: y 8tan θ ( x sec θ) y 6 tanθ x sinθ + tanθ x sinθ + y 0 tanθ Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 b y 0: xsinθ 0tanθ x 0secθ x 0: y 0tanθ y 0tanθ So A has coordinates (0sec θ,0) and B has coordinates (0,0tan θ ) Midpoint of AB has coordinates (0sec θ,tan θ ), so x 0secθ and y tanθ Using sec θ tan θ : 0 The locus of the midpoint is, which is the equation of a hyperbola. 00 cos a Use the chain rule to find the gradient of the tangent: d b t t asint dt asint The gradient of the normal is bcost asint An equation for the line is: y bsin t ( x acos t) bcost bycost b sintcost axsint a sintcost b y 0: x 0: ax t by t a b sin cos ( )sin cos t t a b axsin t ( a b )sintcost x cost a a b bycos t ( a b )sintcost y sint b So M has coordinates a b a cos t,0 and N has coordinates a b a b Midpoint of MN has coordinates: cos t, sint a b a b ax x cost cost a a b a b by y sint sint b a b a x b y Using cos t + sin t : ( a b ) + ( a b ) a x + b y ( a b ) The locus described by the midpoint of MN is a x + b y ( a b ) a b 0, sint b Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 a and b Usingb ( e ) a 6 : 9 ( e ) e e The directrices of the ellipse are the lines of equation x ± P is the point (x, y). By definition, for any point P on an ellipse, the ratio of the distance of P from the focus of the ellipse to the distance of P from the directrix is constant, called the eccentricity e. Let M be the point on the directrix x where PS epm Let M be the point on the directrix x where PS epm PM and PM are parallel to the x-axis. PM + x and PM x 0 SoPS + PS epm + epm 0 bcost 6 y bsin t, x acost asint bcost Equation of l is: y bsin t ( x acos t) asint bxcost + aysint ab a l meets the x-axis at M,0, so the area required is the area of the triangle OMP. cost The height of the triangle is the length of the perpendicular from P to the x-axis bsint Area of shaded triangle OMP base height a abtant bsint cost Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0

11 7 The parametric equations for the ellipse are x 6cos θ, y sin θ, 0 θ < Using the chain rule, the gradient of the tangent is d y cos θ sinθ At point P, : 6cos θ, sinθ tanθ θ At P the gradient of the tangent is tanθ So the equation of line l is By substituting y 0, line y ( x ) l meets the x-axis at Q (, 0) Let R(, 0) be the projection of P on the x-axis. 7 Then the area of the triangle PQR is PQ QR ( ) You need to subtract the region of PQR which is contained in the ellipse. x In the first quadrant the ellipse is described by the function y 6 Since the ellipse meets the x-axis at x 6, the area of the region is given by 6 x the integral 6 Solve this with the substitution x 6sin θ : sin θ(6cos θ)dθ 8 cos θ dθ cosθ 8 d θ sinθ So the area of the shaded region is Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

12 8 a and b, so b a ( e ) e e 9 a 9 Directrices are at ± so they are x ± e Let P and Q be on the right-hand side of the hyperbola. The tangents at P and Q meet the directrix x at A (y 0) and the directrix x at B and C. The point D is where the directrix x crosses the x-axis. The base of triangle ABC is the distance BC, while the height is the distance between the directrices. The parametric equation for H is x sec t, y tant sec Use the chain rule to find the gradient: d t t tant dt sect So the tangent to H has the equation: y tan t ( x sec t) tant y tant tan t x sect sec t tan + (sec tan ) sec y t t t x t ytant + xsect xsect ytant 9 Tangents meet at A, which is,0, so let 9 x, y 0 6 sec t so cost t ± 0.97 sec ± 0.97, tan ± 0.97 ± sect The gradient of the tangent with positive slope is tant BD 8 B is on the tangent, so BD 6 AD 8 08 By symmetry, BD is half of BC, so the area of ABC is BD AD 6 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

13 9 a The parametric equation of the hyperbola H is x sec t, y tant sec Use the chain rule to find the gradient: d t t tant dt sect So the tangent to H has the equation: y tan t ( x sec t) tant ytant tan t xsect sec t ytan t + (sec t tan t) xsect ytant + xsect xsect ytant Tangents meet at,0, so let x, y 0 sec t so sect Using sec t tan t, tan t 8 tant ± So P and Q are (, ) and (, ) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

14 9 b Let A be the point where PQ meets the x-axis, and B be the point where the tangents cross. The area of the triangle ABP is 8 In the first Cartesian quadrant, and for y >, the hyperbola can be seen as the graph of the function y x. This can be integrated: the integral x can be solved by substituting x coshu, as follows: arcosh arcosh sinh udu 0 0 cosh u du arcosh sinh u + 0 arcosh sinh arcosh + arcosh + The area of the shaded region is twice the area of the triangle minus twice the value of the integral, 6 so it is + arcosh 6 arcosh arcosh Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

15 Challenge Let P ( acos θ, bsin θ) The focus has coordinates (ae, 0), so the distance PS ( ae acos θ) + b sin θ PS is: a e a ecosθ + a cos θ + b sin θ The equation of the normal is axsinθ bycos θ ( a b )sinθcosθ ( a b )cosθ This intersects the x-axis at x ae a Then QS ( ae cos θ ae) cosθ a e cos θ a e cosθ + a e QS If QS eps, then QS e PS PS e QS a e cos θ a e cosθ + a e a e cos θ a ecosθ + a e e QS Set PS : e a e cos θ a ecosθ + a a e a ecosθ + a cos θ + b sin θ a e cos θ + a a e + a cos θ + b sin θ Use b a ( e ) a e a b : ( a b )cos θ + a ae + a cos θ + b sin θ a b cos θ ae + b sin θ a ae + b The last equation is true as it is a rearrangement of the defining equation for eccentricity, so this is proved. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2013 HSC Mathematics Extension 2 Marking Guidelines

2013 HSC Mathematics Extension 2 Marking Guidelines 3 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer B A 3 D 4 A 5 B 6 D 7 C 8 C 9 B A 3 HSC Mathematics Extension Marking Guidelines Section II Question