Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

If \mathrm{x+y=\frac{a}{4}+\frac{b}{2}} bisects the 2 distinct chords of \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=2 }which passes through (-a, b)[a, b>0], then \mathrm{ \left|\frac{a}{b}\right|<4 \sqrt{\frac{1}{\lambda}},} then the value of \mathrm{\lambda} is

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=2}             ....[1]

\mathrm{x+y=\frac{a}{4}+\frac{b}{2}}            ...[2]

Let PQ and PR be 2 distinct chords through P(–a, b) and

\mathrm{\text { midpoint of } P Q \text { (or } P R \text { ) be } M\left(t, \frac{a}{4}+\frac{b}{2}-t\right)}         [M lies on (2)]

Let coordinates of Q be (x,y)

\mathrm{\begin{aligned} & \therefore t=\frac{x-a}{2}, \frac{a}{4}+\frac{b}{2}-t=\frac{y+b}{2} \Rightarrow x=2 t+a \& y=\frac{a}{2}-2 t \\ & \therefore Q(x, y) \text { lies on }(1) \Rightarrow \frac{(2 t+a)^2}{a^2}+\frac{\left(\frac{a}{2}-2 t\right)^2}{b^2}=2 \\ & \Rightarrow\left(4 a^2+4 b^2\right) t^2+\left(4 a b^2-2 a^3\right) t+\left(\frac{a^4}{4}-a^2 b^2\right)=0 \end{aligned}}

For 2 distinct roots, roots of this equation must be real and distinct

\mathrm{\begin{aligned} & \Rightarrow\left(4 a b^2-2 a^3\right)^2>4 \cdot\left(4 a^2+4 b^2\right)\left(\frac{a^4}{4}-a^2 b^2\right) \\ & \Rightarrow\left(2 b^2-a^2\right)^2>\left(a^2+b^2\right)\left(a^2-4 b^2\right) \\ & \Rightarrow 8 b^4>a^2 b^2 \Rightarrow\left|\frac{a}{b}\right|<2 \sqrt{2} \end{aligned}}

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question