Question

A man running around a racecourse notes that the sum of the distances of two flag-posts from him is always $10m$ and the distance between the flag-posts is $8m$. Find the area if the path he encloses is
A. $18\pi$ square meters
B. $15\pi$square meters
C. $12\pi$ square meters
D. $8\pi$ square meters

Answer 1

Hint: If we clearly understand the question, we will get to know that the locus of the given question is Ellipse. The two flag posts are represented as two poles of the ellipse. And the man is running on the locus of the ellipse.

Complete step-by-step answer:

Let us note down the given data firstly,

When the man runs around the racecourse, the sum of the distances of two flag-posts from him is always $10m$. The distance between the flag-posts is $8m$.

So, we can relate this situation with the locus of the ellipse.

As per the properties of ellipse,

The sum of the distances to a point on an ellipse from its poles is always constant and that is equal to the length of the Major axis.

From the diagram,

We can say that the sum of the distances between the foci and a point on the ellipse will be equal to $2a$. Apply that in the above diagram, we will get as follows:

${F_1}P + {F_2}P = 2a$

$a$ is the length of the semi-major axis.

It is also given that the sum of the distances of two flag-posts from him is always $10m$.

${F_1}P + {F_2}P = 10$

So, from the above two equations when we compare the right-hand side of the equations, we can conclude that

$2a = 10$

$a = 5$

Now,

The distance between the flag-posts$= 2c$ (From diagram)

But as per the given data, in question it is given $8$m.

So, let us equate them,

$2c = 8$

$c = 4$

See here, now we need to get the value of $b$ in order to calculate the area of the ellipse. So, to find the value of $b$ we can use the above triangle.

We all are greatly familiar with Pythagoras theorem which states that the square of hypotenuse of a right-angled triangle is always equal to the sum of the squares of the remaining sides of the triangle.

If $\vartriangle ABC$ is right-angled at $B$ then we can use Pythagoras theorem as.

${(AC)^2} = {(AB)^2} + {(BC)^2}$

Here $AC$is hypotenuse,$AB$ and $BC$are perpendicular and base of the triangle.

From the diagram, by applying Pythagoras theorem in the $\vartriangle {F_1}OP$

${({F_1}P)^2} = {({F_1}D)^2} + {(DP)^2}$

As we can see in the diagram that ${F_1}P = a,{F_1}D = c$and $DP = b$

So, by substituting the above values we can write this equation as

${a^2} = {c^2} + {b^2}$

Now, we need to find out the value of $b$ hence put it at the left-hand side,

$\Rightarrow {b^2} = {a^2} - {c^2}$

We know the values as $a = 5$ and $c = 4$

So, substitute them in the above equation to simplify further.

${b^2} = {5^2} - {4^2}$

${b^2} = 25 - 16$

${b^2} = 9$

$b = \sqrt 9$

Now we know that the value of $\sqrt 9 = 3$, hence

$b = 3$

At last, we know that the area of the ellipse is $\pi ab$ and we have already calculated a and b that are $a = 5$ and $b = 3$. So, substituting these values in this formula we will get the area of ellipse as follows:

Area of the ellipse $= \pi ab$square meters

$= \pi \times 5 \times 3$ square meters

$= 15\pi$ square meters

So, the required answer for the given problem is $15$ square meters (Option B).

Note: When you get the geometry question in which it states that the sum of distances from a single point remains constant, then most probably it is the locus of the ellipse and that constant distance always must equal to the length of the major axis of the ellipse.