Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The equation of a tangent to the hyperbola $4x^{2}-5y^{2}=20$ parallel to the line $x-y = 2$ is:
Option 1)

$x-y+1=0$
Option 2)

$x-y+7 = 0$
Option 3)

$x-y+9 =0$
Option 4)

$x-y-3 = 0$

Answers (1)

best_answer

Parallel lines -

$\frac{A_{1}}{A_{2}}=\frac{B_{1}}{B_{2}}\neq \frac{C_{1}}{C_{2}}$

- wherein

The two lines are $A_{1}x+B_{1}y+C_{1}=0$ and $A_{2}x+B_{2}y+C_{2}=0$

Condition for Tangency in Hyperbola -

$C^{2}= a^{2}m^{2}-b^{2}$

- wherein

For the Hyperbola

$\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1$

Equation of Tangent to Hyperbola -

$\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}= 1$

- wherein

For the Hyperbola

$\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1$ and

$P\left ( x_{1} ,y_{1}\right )$

from the concept

standard hyperbola

$\frac{x^{2}}{5}-\frac{y^{2}}{4}=1$

slope of tangent = 1

Eqn of tangent $y=x\pm \sqrt{5-4}$

$=>y=x\pm 1$

$=>y=x+ 1 \: \: or\: \: y=x-1$

Option 1)

$x-y+1=0$
Option 2)

$x-y+7 = 0$

Option 3)

$x-y+9 =0$

Option 4)

$x-y-3 = 0$

Posted by

admin

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 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

JEE Mains: Farmer's son from remote Maharashtra village aces exam

Latest Question