Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the angle between pair of tangents drawn to the ellipse <img alt="\mathrm{3 x^2+2 y^2=5}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%20x%5E2&plus;2%20y%5E2%3D5%7D"

Find the angle between pair of tangents drawn to the ellipse \mathrm{3 x^2+2 y^2=5} from the point (1,2).

Option: 1

\mathrm{\tan ^{-1} \frac{\sqrt{12}}{5}}


Option: 2

\mathrm{\tan ^{-1} \frac{12}{\sqrt{5}}}


Option: 3

\mathrm{\tan ^{-1} \frac{5}{3}}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

\mathrm{ \begin{aligned} & 3 x^2+2 y^2=5 \Rightarrow \frac{x^2}{(5 / 3)}+\frac{y^2}{(5 / 2)}=1 \\ & a^2=\frac{5}{3}, b^2=\frac{5}{2} \end{aligned} }

P is the point (1,2)

Equation of pair of tangents

\mathrm{ \begin{aligned} & S S_1=T^2 \\\\ & \left(3 x^2+2 y^2-5\right)(3+2.4-5)=(3(x .1)+2(y \cdot 2)-5)^2 \\\\ & \left(3 x^2+2 y^2-5\right)(6)=(3 x+4 y-5)^2 \\\\ & 18 x^2+12 y^2-30=9 x^2+12 x y-15 x+12 x y+16 y^2-20 y-15 x-20 y+25 \\\\ & 9 x^2-4 y^2-24 x y+30 x+40 y-55=0 \\\\ & \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b}=\frac{2 \sqrt{144+36}}{5}=\frac{2}{5} \sqrt{180}=\frac{12 \sqrt{5}}{5} . \end{aligned} }

Posted by

Gaurav

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

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