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.
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.
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.
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.
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 + 6+ 9sin 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.
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
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
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
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
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
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 6 6 + cosθ 8 d θ 6 9 + 9 sinθ 9 + 9 So the area of the shaded region is 7 + 9 9 6 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
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 9 9 9 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.
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.
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.
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.