Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Tangents drawn from the point (c, d) to the hyperbola $\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$ make angles $\mathrm{ \alpha}$ and $\mathrm{\beta}$ with the x-axis. If $\mathrm{\tan \alpha \tan \beta=1}$ then find the value of $\mathrm{ c^2-d^2.}$
Option: 1
$\mathrm{a^2-b^2}$

Option: 2
$\mathrm{(a b)^2}$

Option: 3
$\mathrm{a-b}$

Option: 4
$\mathrm {a^2+b^2}$

Answers (1)

best_answer

One of the equations of tangents to the hyperbola having slope m is $\mathrm{y=m x+\sqrt{a^2 m^2-b^2}}$ . It passes through (c, d). So,

$\mathrm{ \begin{aligned} & d=m c+\sqrt{a^2 m^2-b^2} \\\\ & \text { or }(d-m c)^2=a^2 m^2-b^2 \\\\ & \Rightarrow\left(c^2-a^2\right) m^2-2 c d m+d^2+b^2=0 \end{aligned} }$

or $\mathrm{(d-m c)^2=a^2 m^2-b^2}$

$\mathrm{ \Rightarrow\left(c^2-a^2\right) m^2-2 c d m+d^2+b^2=0 }$
or Product of roots $\mathrm{ =m_1 m_2=\frac{d^2+b^2}{c^2-a^2}}$

$\mathrm{ \Rightarrow \tan \alpha \tan \beta=\frac{d^2+b^2}{c^2-a^2}=1 \Rightarrow d^2+b^2=c^2-a^2 }$

or $\mathrm{c^2-d^2=a^2+b^2}$

Posted by

Irshad Anwar

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

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question