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Solve the Problem 52ft52 \mathrm{ft} And the Vertical Clearance Must Be 26ft26 \mathrm{ft}

Question 190

Multiple Choice

Solve the problem.
-A railroad tunnel has the shape of half an ellipse. The height of the tunnel at the center is 52ft52 \mathrm{ft} and the vertical clearance must be 26ft26 \mathrm{ft} at a point 24ft24 \mathrm{ft} from the center. Find an equation for the ellipse, where x\mathrm{x} and y\mathrm{y} are measured in ft\mathrm{ft} .
 Solve the problem. -A railroad tunnel has the shape of half an ellipse. The height of the tunnel at the center is  52 \mathrm{ft}  and the vertical clearance must be  26 \mathrm{ft}  at a point  24 \mathrm{ft}  from the center. Find an equation for the ellipse, where  \mathrm{x}  and  \mathrm{y}  are measured in  \mathrm{ft} .     A)   \frac{x^{2}}{768}+\frac{y^{2}}{676}=1   B)   \frac{x^{2}}{576}+\frac{y^{2}}{2704}=1   C)   \frac{x^{2}}{2704}+\frac{y^{2}}{768}=1   D)   \frac{x^{2}}{768}+\frac{y^{2}}{2704}=1


A) x2768+y2676=1\frac{x^{2}}{768}+\frac{y^{2}}{676}=1

B) x2576+y22704=1\frac{x^{2}}{576}+\frac{y^{2}}{2704}=1

C) x22704+y2768=1\frac{x^{2}}{2704}+\frac{y^{2}}{768}=1

D) x2768+y22704=1\frac{x^{2}}{768}+\frac{y^{2}}{2704}=1

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