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Chapter 3 47

# Chapter 3 47 - SECTION3.6 C 197 63 To ﬁnd the points at...

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Unformatted text preview: SECTION3.6 IMPLICITDIFFERENTIATION C 197 63. To ﬁnd the points at which the ellipse 21:2 — my + y2 2 3 crosses the x-axis. let y 2 0 and solve for x. 64. 65. 66. y 2 0 2 x2 — 33(0) + 02 2 3 42 a: 2 213/3. So the graph of the ellipse crosses the x—axis at the points (ix/3, 0). Using implicit differentiation to ﬁnd 7/, we get 2x ~ my’ — y + 2yy’ 2 0 2> y’(2y — 3:) 2 y — 2x y—2a: , , 022t/Ei , _ 0+2\/§_ <=> ' 2 . So at 3,0 1s ———-— — 2 and at ~\/3,0 13 —~— — 2. Thus. the tangent y 2y—ar y (f) 2(0)2\/§ y ( ) 2(0)+\/§ lines at these points are parallel. . . . . . . , y — 2x . , (a) We use 1mp11c1td1fferentiat10n to ﬁnd y 2 2y _ 30 as in Exerc1se 63. (b) . . 1 — 2(—1) 3 1 T 1 : —— : — : ]_a The slope ofthe tangent me at ( 1, )1s m 2(1) _(_1) 3 1 . so the slope of the normal line is —— 2 —1. and its equation 15 m y — 1 2 —1(:c +1) (2 y 2 —\$. Substituting —:n fory in the equation of the ellipse, we get :52 — m(—93) + (—33)2 = 3 g 39:2 2 3 42 a: 2 :1. So the normal line must intersect the ellipse again at m 2 1, and since the equation of the line is y 2 —x, the other point of intersection must be (1, —1). w2y2+wy22 2> x2-2yy’+y2-2m+a:-yl+y-12O 42> y’(2m2y+m)2—2my2—y (2 25392 -- y 2903/2 -— y lﬁ _ 2 g 2 g y 2\$2y——x'so 2\$2y--\$_ 1 290g +y—2my+\$ <=> y(2a:y+1)—w(2a:y+1) (2) 31(293y+1)—\$(2xy+1):0 <2 (2my+1)(y—33)20 <2 \$y2—%ory2x.Buta:y27% 2> x2y2+\$y2i—%22,sowemusthavem2y.Then w2y2—l—ary22 => \$4+\$222 <2 m4 + m2 — 2 2 0 (2 (m2 —|— 2) (\$2 — 1) 2 0. So 902 2 —2, which is impossible, or 3:2 2 1 (2 a: 2 :1. Since a: 2 y. the points on the curve where the tangent line has a slope of —1 are (21, —1) and (1, 1). :52 + 4312 2 36 2> 293 + 8yy’ 2 0 2> y’ 2 —%. Let (a, b) be a point on m2 + 43/2 2 36 whose tangent line passes through (12, 3). The tangent line is then y — 3 2 _41b (x — 12), so b 2 3 : 24gb both sides by 4b gives 4b2 — 12b 2 —a2 + 12a, so 11b2 + a2 2 12(a + b). But 4b2 + a2 2 36, so 36 2 12(a + b) 2 a + b 2 3 2> b 2 3 — 0.. Substituting 3 — a for b into a2 + 4b2 2 36 gives a2 + 4(3 — a)2 2 36 <=> a2+36—24a+4a2236 4: 5(12 ~24a20 42> a(5a~24) 20.soa200ra2 2. Ifa20, (a — 12). Multiplying 5 b I 3 — 0 = 39 and ifa = %, b 2 3 — % 2 “g. So the twopoints on the ellipse are (0,3) and (%, —§). Using y — 3 2 2%(w — 12) with (a, b) 2 (0,3) gives us the tangent liney — 3 2 Oory 2 3. With (a, b) 2 (%, —§), we have 4 y—32—M—2_g%(m—12) 42> y—3:§(z—12) 42> y 2 gm — 5. A graph of the ellipse and the tangent lines conﬁrms our results. ...
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