Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the radii of director circle of the ellipse <img alt="\frac{x^2}{a^2}+\frac{y^2}{b^2}=1" src="https://learn.careers360.com/latex-image/?%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac%7By%5E

If the radii of director circle of the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and hyperbola  \frac{x^2}{a^2}-\frac{y^2}{b_1^2}=1 are in the ratio 1:2 and 4e_1^2-e^2_2=\lambda where e1 and e2 are eccentricities of ellipse and hyperbola respectively, then the value of \lambda is  

Option: 1

2


Option: 2

4


Option: 3

6


Option: 4

8


Answers (1)

best_answer

 

 

Director Circle:

\\\text {The equation of the director circle of the hyperbola } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \text { is } \\\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}-\mathrm{b}^{2}

The locus of the point through which perpendicular tangents are drawn to a given Hyperbola S = 0 is a circle called the director circle of the hyperbola.


Director Circle of Hyperbola:

\\ {\text {Equation of tangent of the hyperbola } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { in slope form is } y=m x+\sqrt{a^{2} m^{2}-b^{2}}} \\ {\text { it passes through the point }(h, k)} \\ {k=m h+\sqrt{a^{2} m^{2}-b^{2}}} \\ {(k-m h)^{2}=a^{2} m^{2}-b^{2}} \\ {k^{2}+m^{2} h^{2}-2 m h k=a^{2} m^{2}-b^{2}} \\ {\left(h^{2}-a^{2}\right) m^{2}-2 h k m+k^{2}+b^{2}=0} \\ {\text { This is quadratic equation in m, slope of two tangents are } m_{1} \text { and } m_{2}} \\ {\mathrm{m}_{1} \mathrm{m}_{2}=\frac{\mathrm{k}^{2}+\mathrm{b}^{2}}{\mathrm{h}^{2}-\mathrm{a}^{2}}} \\\\ -1=\frac{\mathrm{k}^{2}+\mathrm{b}^{2}}{\mathrm{h}^{2}-\mathrm{a}^{2}} \\\\ {\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}-\mathrm{b}^{2}}

 

-

 

 

Director circle of the ellipse is x^2+y^2=a^2+b^2

\therefore \;\;\;\;\;\; \text{radius }r_1=\sqrt{a^2+b^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;=\sqrt{a^2+a^2(1-e_1^2)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;=a\sqrt{2-e_1^2}

Director circle of the hyperbola is x^2+y^2=a^2-b^2_1

\therefore \;\;\;\;\;\; \text{radius }r_2=\sqrt{a^2-b_1^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;=\sqrt{a^2-a^2(e_2^2-1)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;=a\sqrt{2-e_2^2}

given,

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\frac{r_1}{r_2}=\frac{1}{2}\\\\\Rightarrow \;\;\;\;\;\;\frac{a\sqrt{2-e_1^2}}{a\sqrt{2-e_2^2}}=\frac{1}{2}\\\\\Rightarrow \;\;\;\;\;\;\frac{{2-e_1^2}}{{2-e_2^2}}=\frac{1}{4}\\\\\Rightarrow \;\;\;\;\;\;8-4e^2_1=2-e^2_2\\\\\Rightarrow \;\;\;\;\;\;4e^2_1-e^2_2=6=\lambda\\\lambda = 6

 

Posted by

seema garhwal

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 withheld results declared; NTA issues show cause notices ahead of action
anam-khan

JCECEB 2025 registration for BTech admission through JEE Main rank begins jceceb.jharkhand.gov.in
anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective

Latest Question