Question

A stair–case of length l rests against a vertical wall and a floor of a room. Let P be a point on the stair-case, nearer to its end on the wall that divides its length in the ratio 1:2.If the stair-case begins to slide on the floor, then the locus of P is:

Answer 1

Hint- The given terms will form a shape of triangle and
Formula used:
Section formula:
 The coordinate of P(x,y) which divides the line segment joining the points A\[({x_1},{y_1})\] and B\[({x_2},{y_2})\]internally in the ratio m:n are
\[(\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}})\]

Complete step-by-step answer:
It is given that, a stair–case of length l rests against a vertical wall and a floor of a room.

Let, “b” be the height and “a” be the length intercepted by the stair-case.
If we consider BC as x axis and BA as y axis and B as the origin (0,0) point ,then the coordinate of C is (a,0) and the coordinate of A is (0,b).
Now, P be a point on the stair-case, nearer to its end on the wall that divides its length in the ratio 1:2.
Then we can consider the section formula to calculate the coordinate of P.
The coordinate of P(x,y) which divides the line segment joining the points A\[(0,b)\] and C\[(a,0)\]internally in the ratio 1:2 are
\[\left( {\dfrac{{1 \times a + 2 \times 0}}{{1 + 2}},\dfrac{{1 \times 0 + 2 \times b}}{{1 + 2}}} \right)\]
By solving the above points we get,
\[\left( {\dfrac{a}{3},\dfrac{{2b}}{3}} \right)\]
Now, the length of the stair-case l is constant.
From the triangle ABC by Pythagoras theorem, we have
\[{a^2} + {b^2} = {l^2}\] …..(1)
Let us consider, the coordinate of P is (x,y)
Then,
\[x = \dfrac{a}{3}\& y = \dfrac{{2b}}{3}\]
Let us find a and b from the above equation we get,
\[a = 3x\& b = \dfrac{{3y}}{2}\]
Putting the values in (1) we get,
\[{(3x)^2} + {(\dfrac{{3y}}{2})^2} = {l^2}\]
On solving we get,
\[9{x^2} + \dfrac{{9{y^2}}}{4} = {l^2}\]
Let us rewrite the above equation as follows,
\[\dfrac{{{x^2}}}{{\dfrac{{{l^2}}}{9}}} + \dfrac{{{y^2}}}{{\dfrac{{4{l^2}}}{9}}} = 1\]
This represents the equation of an ellipse.
The general equation of ellipse is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] and the locus is given by \[1 - {e^2} = \dfrac{{{a^2}}}{{{b^2}}},{b^2} > {a^2}\]
So, locus of P is
\[e = \sqrt {1 - \dfrac{{\dfrac{{{l^2}}}{9}}}{{\dfrac{{4{l^2}}}{9}}}} \]
Let us solve the above equation as follows,
\[e = \sqrt {1 - \dfrac{{{l^2}}}{9} \times \dfrac{9}{{4{l^2}}}} \]
On final solving we get
\[e = \sqrt {\dfrac{{4 - 1}}{4}} = \dfrac{{\sqrt 3 }}{2}\]
Hence, locus of P is \[\dfrac{{\sqrt 3 }}{2}\]

Note: Pythagoras theorem used in the problem states that in a triangle the sum of the square of the opposite and adjacent sides are equal to the square of hypotenuse.
The locus of the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] is given by \[1 - {e^2} = \dfrac{{{a^2}}}{{{b^2}}},{b^2} > {a^2}\]