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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 :

Option 1)
an ellipse of eccentricity $\frac{1}{2}\;$

Option 2)
an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

Option 3)
a circle of radius $\frac{l}{2}$

Option 4)
a circle of radius $\frac{\sqrt{3}}{2}l$

Answers (1)

best_answer

As we learnt in

Selection formula -

$x= \frac{mx_{2}+nx_{1}}{m+n}$

$y= \frac{my_{2}+ny_{1}}{m+n}$

- wherein

If P(x,y) divides the line joining A(x₁,y₁) and B(x₂,y₂) in ration $m:n$

and

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

We know length = l

so, x², + y² = l²

now using section formula

P(h, k) is $(\frac{x}{3},\frac{2y}{3})$

$so, h=\frac{x}{3}; k=\frac{2y}{3}$

x = 3h; $y=\frac{3}{2}k$

$9h^{2}+\frac{9k^{2}}{4}=l^{2}$ , an ellipse

$\frac{h^{2}}{\frac{l^{2}}{9}}+\frac{k^{2}}{\frac{4l^{2}}{9}}=1$

so $e=\sqrt{\frac{1-l^2}{9\times \frac{4l^2}{9}}} = \frac{\sqrt{3}}{2}$

Option 1)

an ellipse of eccentricity $\frac{1}{2}\;$

this is incorrect

Option 2)

an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

this is correct

Option 3)

a circle of radius $\frac{l}{2}$

this is incorrect

Option 4)

a circle of radius $\frac{\sqrt{3}}{2}l$

this is incorrect

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