Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let $\mathrm {a}$ and $\mathrm {b}$ respectively be the semi-transverse and semiconjugate axes of a hyperbola whose eccentricity satisfies the equation $\mathrm {9 e^2-18 e+5=0}$ . If $\mathrm {S(5,0)}$ is a focus and $\mathrm {5 x=9}$ is the corresponding directrix of this hyperbola, then $\mathrm {a^2-b^2}$ is equal to

Option: 1
$-7$

Option: 2
$-5$

Option: 3
$5$

Option: 4
$7$

Answers (1)

best_answer

$S(5,0)$ is the focus $\mathrm { \therefore a e=5 ......(i)}$

$\mathrm { 5 x=9\: \: i.e., x=\frac{9}{5} \: \: is \: \: the\: \: directrix \therefore \frac{a}{e}=\frac{9}{5}......(ii) }$

From (i) and (ii), we get $\mathrm { a=3 \: \: and\: \: e=\frac{5}{3} }$

Now, $\mathrm {b^2=a^2\left(e^2-1\right)=9\left(\frac{25}{9}-1\right)=16 }$

Hence, $\mathrm {a^2-b^2=9-16=-7 }$

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

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions

anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question