Question

a wooden plaque is in the shape of an ellipse with height 26 centimeters and width 16 centimeters. find an equation for the ellipse and use it to find the horizontal width, in centimeters, of the plaque at a distance of 6 centimeters above the center point. (round your answer to the nearest hundredth if necessary.) provide your answer below: cm

Explanation:

Step1: Determine the semi - axes

The standard form of an ellipse centered at the origin is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$. The width is $16$ cm, so the semi - major axis $a=\frac{16}{2}=8$ cm (assuming the ellipse is wider horizontally), and the height is $26$ cm, so the semi - minor axis $b = \frac{26}{2}=13$ cm. The equation of the ellipse is $\frac{x^{2}}{64}+\frac{y^{2}}{169}=1$.

Step2: Substitute the $y$ - value

We want to find the horizontal width at $y = 6$ cm above the center. Substitute $y = 6$ into the equation of the ellipse: $\frac{x^{2}}{64}+\frac{6^{2}}{169}=1$. Then $\frac{x^{2}}{64}=1-\frac{36}{169}=\frac{169 - 36}{169}=\frac{133}{169}$.

Step3: Solve for $x$

Cross - multiply to get $169x^{2}=133\times64$. Then $x^{2}=\frac{133\times64}{169}$, and $x=\sqrt{\frac{133\times64}{169}}$. Since the width is $2|x|$, we have $2x = 2\sqrt{\frac{133\times64}{169}}$.
$2x=2\sqrt{\frac{8512}{169}}\approx2\sqrt{50.3669}\approx2\times7.0976\approx14.1952\approx14.20$ cm.

Answer:

$14.20$