2 b. 3 Use the chain rule to find the gradient:

Transcription

1 Conic sections D x cos θ, y sinθ d y sinθ So tngent is y sin θ ( x cos θ) sinθ Eqution of tngent is x + y sinθ sinθ Norml grdient is sinθ So norml is y sin θ ( x cos θ) xsinθ ycos θ ( )sinθ, So eqution of tngent is: x + ysinθ xsinθ y sinθ + 5, 5 9 So eqution of tngent is: x + 5ysinθ 5 5xsinθ y sinθ x d + y x + y y 9 9 d x x 5 so t 5, m 9y Tngent t 5 5, is y ( x 5) y+ 5x 9 Norml t 5, is y ( x 5) 5 5y 5 8x 8 5 5y 8x x so t (, ) m y Tngent t (, )is y ( x ( )) y x 8 Norml t (, )isy ( x ( )) y+ x Use the chin rule to find the grdient: d cost t Then the eqution of the tngent is given y cost ( y sin t) ( x cos t) y sin t xcost+ cos t xcost+ y Using the method in Exmple, sustitute for y in the eqution of the ellipse. x x + y + ( x+ 5) So x + ( x + 5x+ 5) 5x + 8 5x+ This hs discriminnt: ( 8 5) 5 So the line meets the ellipse t only one point nd therefore is tngent to the ellipse. Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.

2 To find the point of contct, solve the eqution from prt for x: 5x + 8 5x+ 5 x+ ( 5 ) x 5 y So the point of contct is 5, 5 x cos θ, y sinθ sinθ sinθ So grdientofnormlis cos θ sinθ y sin θ ( x cos θ) y sinθ xsinθ 9sinθ y xsinθ 5sinθ 5 Norml t P crosses the x-xis t 5 y, x Sustituting into the eqution for the norml from prt : 5 sin θ 5sin θ cos θ sinθ + sinθ or sinθ gives θ or 8 gives θ or θ x, y θ 8 x, y θ x, y θ x, y So the coordintes of other possile positions of P re (,),(,), or, y mx+ c is tngent to if m + c y x+ c m, c? y x +, m + c + c c 8 so c± + Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.

3 7 Sustitute y mx + into the eqution for the ellipse. ( mx+ ) x x + ( mx+ ) 5 (5 + m ) x + mx+ Since the line is tngent the discriminnt of this eqution must equl zero (must hve equl roots). So m (5 + m ) 8+ m m 8 m m± 8 c Grdient of norml is Eqution of norml t P is y x x y So A is (, ) d Tngent is y x+, x y The m + c condition could 8 e used s in question. y mx+, + c,, Using the condition m + c + m m m ± But m >, so m m + c : y x+, + Sustitute into the eqution for the ellipse: x (x + x+ ) + x + x + x+ x + x+ 9 (x+ ) x, y x+ So P is (,) 9 So B is (, ) Are of APB 5 5 cos θ, sinθ dθ dθ cot θ sinθ 9 8 ( ) ( ) RHS So Q, lies on E c Let Q e the point (cos φ,sin φ ) Using the coordintes of Q: 9 cosφ cosφ 8 sinφ sinφ So cotφ Grdient of tngent t Q is cotφ Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.

4 9 d Tngent t P is perpendiculr to tngent t Q, so grdient of tngent t P cotθ tnθ 9 So P is, Use the chin rule to find the grdient of the tngent: 8sinθ sinθ The eqution of the tngent l is given y y sin θ ( x 8cos θ) sinθ θ θ x θ + θ ysin 8sin cos 8cos or 9 P is, Eqution of the tngent is x + ysinθ 8 Grdient of the norml l is sin θ nd its eqution is sinθ y sin θ ( x 8cos θ) y sinθ xsinθ sinθ y mx+ c nd + 9 9, + m c + 9m c y mx+ c m c + 5m c () m nd 5, () () m m nd + 5m c c c± 8 So m±, c± 8 m± () Eqution of the norml is: x sinθ y sinθ Line l cuts the x-xis t A, so y : x 8 so x 8secθ A is the point (8sec θ,) Line l cuts the y-xis t B, so x : y sinθ so y sinθ B is the point (, sin θ) Now find the eqution of the line AB. y ( sin θ) x 8secθ 8secθ y x 8secθ sinθ 8secθ ysecθ xsinθ secθsinθ xsinθ ysecθ secθsinθ xsinθ y sinθ Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.

5 Use the chin rule to find the grdient: d θ 5sinθ dθ Then the eqution ofl is given y y sin θ ( x 5cos θ) 5sinθ 5ysinθ 5sin θ x + 5cos θ x + 5ysinθ 5 At Q the line cuts the y-xis, so x Sustitute in the eqution for line l: 5ysinθ 5 so y sinθ The point Q hs coordintes, sinθ The grdient of ny line perpendiculr to l is: 5sinθ Then the eqution of l is: 5sinθ y x sinθ ysinθ 9 5xsin θ θ θ θ θ 5xsin ysin cos 9cos c If lcuts the x-xis t (, ), then sustituting into the eqution for l gives θ sin θ 9 ( cos ) 9cos θ cos 9cos cos θ + 9 Using the qudrtic formul to solve gives: θ θ Use the chin rule to find the grdient: d cost cost t Then n eqution for l is given y: cost y sin t ( x cos t) y sin t xcost+ cos t xcost+ y Since l is perpendiculr to l, the grdient of l is cost The line psses through the origin, so the eqution is y mx y x cost Sustituting y x into the eqution cost of l to find the coordintes of the intersection gives: sin t xcost+ x cost xcos t+ xsin t 8cost 8cost x cos t+ sin t Then 8cost y cos cos + sin cos t+ sin t t t t The coordintes of Q re: 8cost, cos t+ sin t cos t+ sin t 9± 8 ( ) 9± 5 Oviously <, so 5 Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 5

6 Use the chin rule to find the grdient: d cost t The eqution of the tngent is: cost y sin t ( x cos t) y sin t xcost+ cos t xcost+ y To find the x-intercept, sustitute y in the eqution for the tngent: xcost x cost To find the y-intercept, sustitute x in the eqution for the tngent: y y The re of the shded tringle is: cost cost cosec t 5 Rerrnging the eqution for the ellipse: y x x y In the first Crtesin qudrnt, the ellipse cn e seen s the grph of the function x y To find the re in the first qudrnt, integrte from x to x The integrl d x xcn e solved y sustituting sinu x, s follows: sin u(cos )d sin u(cos )d cos d u u (+ cos u)du + sin + sin [ u] u u u u Using the identity cos θ + The re of the shded region is twice the vlue of the integrl, so it is 8 Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free.

7 Chllenge In the first Crtesin qudrnt, the ellipse cn e seen s the grph of the functiony x This cn e integrted. The integrl x x sinu, s follows: cn e solved y sustituting sin u( cos u)du cosudu + cosu d u + [ sinu] The re of the ellipse is four times the re contined in single qudrnt, so it is Person Eduction Ltd 8. Copying permitted for purchsing institution only. This mteril is not copyright free. 7