Go To Your Study Dashboard

If the line $y=mx+7\sqrt{3}$ is normal to the hyperbola $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$ , then a value of $m$ is :

Option 1)
$\frac{\sqrt{5}}{2}$

Option 2)
$\frac{\sqrt{15}}{2}$

Option 3)
$\frac{2}{\sqrt{5}}$

Option 4)
$\frac{3}{\sqrt{5}}$

Answers (1)

V Vakul

$y=mx+7\sqrt{3}$

given hyperbola $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$

Normal to hyperbola is slope form

$y=mx\mp \frac{m\left ( a^{2}+b^{2} \right )}{\sqrt{a^{2}-m^{2}b^{2}}}$

compare this

$7\sqrt{3}=\frac{m\left ( 24 +18\right )}{\sqrt{24-18m^{2}}}$

$m=\frac{2}{\sqrt{5}}$

Option 1)

$\frac{\sqrt{5}}{2}$

Option 2)

$\frac{\sqrt{15}}{2}$

Option 3)

$\frac{2}{\sqrt{5}}$

Option 4)

$\frac{3}{\sqrt{5}}$

Similar Questions

Latest Asked Questions

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-

Buy Now

Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-

Buy Now

Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 2999/- ₹ 1999/-

Buy Now

Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 19999/- ₹ 14499/-

Buy Now

Knockout JEE Main (Eight Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 15999/- ₹ 12499/-

Buy Now

Boost your Preparation for JEE Main 2021 with Personlized Coaching

Start Now

Exams

Articles

Questions