Question

Find eccentricity of 16x²+25y²=100

Answer 1

Answer:

$\mathrm{Ellipse\:eccentricity\:given}\:16x^2+25y^2=100:\quad \frac{3}{5}$

Step-by-step explanation:

$\mathrm{Ellipse\:eccentricity}$

• The eccentricity is a measure of how much the ellipse deviates from a circle.
• For an ellipse with major axis parallel to the x-axis, the eccentricity is $\dfrac{\sqrt{a^2-b^2}}{a}$

$=\dfrac{\sqrt{a^2-b^2}}{a}$

$\mathrm{Calculate\:ellipse\:properties}$

$16x^2+25y^2=100$

Ellipse standard equation

• $\dfrac{\left(x-h\right)^2}{a^2}+\dfrac{\left(y-k\right)^2}{b^2}=1\:\mathrm{is\:the\:ellipse\:standard\:equation}$
• $\mathrm{with\:center}\:\left(h,\:k\right)\:\mathrm{and\:}a,\:b\mathrm{\:are\:the\:semi-major\:and\:semi-minor\:axes}$

$\mathrm{Rewrite}\:16x^2+25y^2=100\:\mathrm{in\:the\:form\:of\:the\:standard\:ellipse\:equation}$

$16x^2+25y^2=100$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}16$

$x^2+\dfrac{25}{16}y^2=\dfrac{25}{4}$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}25$

$\dfrac{1}{25}x^2+\dfrac{1}{16}y^2=\dfrac{1}{4}$

$\mathrm{Divide\:by}\:\dfrac{1}{4}$

$\dfrac{x^2}{\dfrac{25}{4}}+\dfrac{y^2}{4}=1$

$\mathrm{Rewrite\:in\:standard\:form}$

$\dfrac{\left(x-0\right)^2}{\left(\dfrac{5}{2}\right)^2}+\dfrac{\left(y-0\right)^2}{2^2}=1$

$\mathrm{Therefore\:ellipse\:properties\:are:}$

$\left(h,\:k\right)=\left(0,\:0\right),\:a=\dfrac{5}{2},\:b=2$

$a>b\:\mathrm{therefore}\:a\:\mathrm{is\:semi-major\:axis\:and}\:b\:\mathrm{is\:semi-minor\:axis}$

$\mathrm{Ellipse\:with\:center}\:\left(h,\:k\right)=\left(0,\:0\right),\:\:\mathrm{semi-major\:axis}\:a=\dfrac{5}{2},\:\:\mathrm{semi-minor\:axis}\:b=2$

$=\dfrac{\sqrt{\left(\dfrac{5}{2}\right)^2-2^2}}{\dfrac{5}{2}}$

$\dfrac{\sqrt{\left(\dfrac{5}{2}\right)^2-2^2}}{\dfrac{5}{2}}$

$2^2=4$

$=\dfrac{\sqrt{\left(\dfrac{5}{2}\right)^2-4}}{\dfrac{5}{2}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{a}{\dfrac{b}{c}}=\dfrac{a\cdot \:c}{b}$

$=\dfrac{\sqrt{\left(\dfrac{5}{2}\right)^2-4}\cdot \:2}{5}$

$\sqrt{\left(\dfrac{5}{2}\right)^2-4}$

$\left(\dfrac{5}{2}\right)^2$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

$=\dfrac{5^2}{2^2}$

$=\dfrac{25}{4}$

$=\sqrt{\dfrac{25}{4}-4}$

$\dfrac{25}{4}-4$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:4=\dfrac{4\cdot \:4}{4}$

$=\dfrac{25}{4}-\dfrac{4\cdot \:4}{4}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{25-4\cdot \:4}{4}$

$25-4\cdot \:4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:4=16$

$=25-16$

$\mathrm{Subtract\:the\:numbers:}\:25-16=9$

$=9$

$=\dfrac{9}{4}$

$=\sqrt{\dfrac{9}{4}}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\dfrac{\sqrt{9}}{\sqrt{4}}$

$=\dfrac{3}{2}$

$\mathrm{Multiply\:}\dfrac{3}{2}\cdot \:2$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}$

$=\dfrac{3\cdot \:2}{2}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=3$

$=\dfrac{3}{5}$

Answer 2

Answer:

5x^33x

Step-by-step explanation: