Question #224006

Write down the parametric equations of the following ellipses. Find their eccentricities, foci and directrices. (a) 4x2 + 9y2 = 4 (b) x2/9 + y2/16 = 1 (c) x2/7 + y2/14 = 1


Expert's answer

(a)

4x2+9y2=44x^2+9y^2=4

x21+y24/9=1\dfrac{x^2}{1}+\dfrac{y^2}{4/9}=1

x=cos⁡t,y=23sin⁡t,0≤t≤2πx=\cos t, y=\dfrac{2}{3}\sin t, 0\leq t\leq 2\pi

r⃗=⟨cos⁡t,23sin⁡t⟩,0≤t≤2π\vec r=\langle\cos t, \dfrac{2}{3}\sin t\rangle, 0\leq t\leq 2\pi

a=1,b=23,c=a2−b2=(1)2−(23)2=53a=1, b=\dfrac{2}{3}, c=\sqrt{a^2-b^2}=\sqrt{(1)^2-(\dfrac{2}{3})^2}=\dfrac{\sqrt{5}}{3}


e=ca=53e=\dfrac{c}{a}=\dfrac{\sqrt{5}}{3}

Foci:(−53,0),(53,0)Foci: (-\dfrac{\sqrt{5}}{3}, 0), (\dfrac{\sqrt{5}}{3}, 0)

Directrices:x=±a2cDirectrices: x=\pm\dfrac{a^2}{c}

first directrix:x=−355,first\ directrix: x=-\dfrac{3\sqrt{5}}{5},

second directrix:x=355second\ directrix:x=\dfrac{3\sqrt{5}}{5}

(b)

x29+y216=1\dfrac{x^2}{9}+\dfrac{y^2}{16}=1

x=3cos⁡t,y=4sin⁡t,0≤t≤2πx=3\cos t, y=4\sin t, 0\leq t\leq 2\pi

r⃗=⟨3cos⁡t,4sin⁡t⟩,0≤t≤2π\vec r=\langle3\cos t, 4\sin t\rangle, 0\leq t\leq 2\pi

a=4,b=3,c=a2−b2=(4)2−(3)2=7a=4, b=3, c=\sqrt{a^2-b^2}=\sqrt{(4)^2-(3)^2}=\sqrt{7}


e=ca=74e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}

Foci:(−7,0),(7,0)Foci: (-\sqrt{7}, 0), (\sqrt{7}, 0)

Directrices:y=±a2cDirectrices: y=\pm\dfrac{a^2}{c}

first directrix:y=−1677,first\ directrix: y=-\dfrac{16\sqrt{7}}{7},

second directrix:y=1677second\ directrix:y=\dfrac{16\sqrt{7}}{7}


(c)

x27+y214=1\dfrac{x^2}{7}+\dfrac{y^2}{14}=1

x=7cos⁡t,y=14sin⁡t,0≤t≤2πx=\sqrt{7}\cos t, y=\sqrt{14}\sin t, 0\leq t\leq 2\pi

r⃗=⟨7cos⁡t,14sin⁡t⟩,0≤t≤2π\vec r=\langle\sqrt{7}\cos t, \sqrt{14}\sin t\rangle, 0\leq t\leq 2\pi

a=14,b=7,c=a2−b2=(14)2−(7)2=7a=\sqrt{14}, b=\sqrt{7}, c=\sqrt{a^2-b^2}=\sqrt{(\sqrt{14})^2-(\sqrt{7})^2}=\sqrt{7}e=ca=22e=\dfrac{c}{a}=\dfrac{\sqrt{2}}{2}

Foci:(−7,0),(7,0)Foci: (-\sqrt{7}, 0), (\sqrt{7}, 0)

Directrices:y=±a2cDirectrices: y=\pm\dfrac{a^2}{c}

first directrix:y=−27,first\ directrix: y=-2\sqrt{7},

second directrix:y=27second\ directrix:y=2\sqrt{7}




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