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 staircase begins to

Correct answer: B. an ellipse of eccentricity 3 2 Let point A(a, 0) is on x -axis and B(0, b) is on y -axis. Let P(h, k) divides AB in the ratio 1: 2 . So, by section formula aligned &h=(2(0)+1(a))/(1+2)=(a)/(3) &k=(2(b)+1(0))/(3)=(2 b)/(3) & a=3 h and b=(3 k)/(2) aligned Now, a^2+b^2=l^2 aligned & 9 h²+(9 k^2)/(4)=l^2 & (h^2)/(((l)/(3))^2)+(k^2)/(((2 l)/(3))^2)=1 aligned Now e= 1-((l^2)/(9) × (9)/(4 l^2)) = 1-(1)/(4) = 3 2 Thus, required locus of P is an ellipse with eccentricity 3 2 .

JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Ellipse

JEE Main 2014 · 11 April · Online · Mathematics · Ellipse

A stair-case of length l rests against a vertical wall and a floor...

A stair-case of length

l

rests against a vertical wall and a floor of a room. Let

\mathrm{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 staircase begins to slide on the floor, then the locus of

\mathrm{P}

is:

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,

Options

Aan ellipse of eccentricity $\frac{1}{2}$
Ban ellipse of eccentricity $\frac{\sqrt{3}}{2}$
Ca circle of radius $\frac{1}{2}$
Da circle of radius $\frac{\sqrt{3}}{2} l$

Correct answer

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

Step-by-step solution

Let point

\mathrm{A}(a, 0)

is on

x

-axis and

\mathrm{B}(0, b)

is on

y

-axis.

Let

\mathrm{P}(h, k)

divides

\mathrm{AB}

in the ratio

1: 2

. So, by section formula

\begin{aligned}&h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3}\\ &k=\frac{2(b)+1(0)}{3}=\frac{2 b}{3}\\ &\Rightarrow a=3 h \text{\text{ and } }b=\frac{3 k}{2} \end{aligned}

Now,

a^2+b^2=l^2

\begin{aligned}&\Rightarrow 9 h²+\frac{9 k^2}{4}=l^2 \\ &\Rightarrow \frac{h^2}{\left(\frac{l}{3}\right)^2}+\frac{k^2}{\left(\frac{2 l}{3}\right)^2}=1 \end{aligned}

Now

\mathrm{e}=\sqrt{1-\left(\frac{l^2}{9} \times \frac{9}{4 l^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}

Thus, required locus of

\mathrm{P}

is an ellipse with eccentricity

\frac{\sqrt{3}}{2}$.

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