A man running round a race-course notes that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The path traced by the man is an ellipse because the sum of the distances from two fixed points (foci) is constant.Let the constant sum be $2a = 1

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) The path traced by the man is an ellipse because the sum of the distances from two fixed points (foci) is constant.Let the constant sum be $2a = 10 \ m$, so $a = 5 \ m$.The distance between the foci is $2ae = 8 \ m$, so $ae = 4 \ m$.Since $a = 5$, we have $5e = 4$, which gives $e = \frac{4}{5}$.Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 25(1 - (\frac{4}{5})^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.Thus, $b = 3 \ m$.The area of the ellipse is given by $\pi ab = \pi 	imes 5 	imes 3 = 15 \pi \ m^2$.

$A$ man running round a race-course notes that the sum of the distances of two flag-posts from him is always $10 \ m$ and the distance between the flag-posts is $8 \ m$. The area of the path he encloses in square metres is (in $\pi$)

Similar Questions

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

One of the foci of an ellipse is $(2,-3)$ and its corresponding directrix is $2x+y=5$. If the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$,then the coordinates of the other focus are

Tangents are drawn from points on the circle $x^2 + y^2 = 49$ to the ellipse $\frac{x^2}{25} + \frac{y^2}{24} = 1$. The angle between the tangents is:

The area of the region bounded by the curve $x = 4 \cos \theta, y = 3 \sin \theta$ is . . . . . . sq. units. (in $\pi$)

The equations of the latus rectum of the ellipse $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module