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Let  \mathrm { A B C}  be an equilateral triangle inscribed in the circle   \mathrm { x^2+y^2=a^2} , suppose perpendiculars from   \mathrm {A, B, C}  to the major axis of the ellipse    \mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ;(a>b)}  meet the ellipse respectively at   \mathrm {P, Q, R }  so that   \mathrm {P, Q, R }  on the same side of the major axis as   \mathrm {A, B, C }  respectively then the normals to the ellipse drawn at the points   \mathrm {P, Q }   and   \mathrm {R } are
 

Option: 1

Non-concurrent
 


Option: 2

Concurrent
 


Option: 3

Can't be said
 


Option: 4

None


Answers (1)

best_answer

Let coordinates of \mathrm {\triangle A B C }  inscribed in circle   \mathrm {x^2+y^2=a^2 }  be  \mathrm {(a \cos \theta, a \sin \theta); } 
\mathrm { \begin{aligned} & \left(a \cos \left(\theta+\frac{2 \pi}{3}\right), a \sin \left(\theta+\frac{2 \pi}{3}\right)\right) \text { and } \\ & \left(a \cos \left(\theta-\frac{2 \pi}{3}\right), a \sin \left(\theta-\frac{2 \pi}{3}\right)\right) \end{aligned} }
then co-ordinates of \mathrm {P, Q, R}  are
\mathrm {(a \cos \theta, b \sin \theta) ;\left(a \cos \left(\theta+\frac{2 \pi}{3}\right), b \sin \left(\theta+\frac{2 \pi}{3}\right)\right) \text { and }}

\left(a \cos \left(\theta-\frac{2 \pi}{3}\right), b \sin \left(\theta-\frac{2 \pi}{3}\right)\right)

Equation of normal at \mathrm {P, Q, R}  to ellipse

\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2 ......(1)\\ & \frac{a x}{\cos \left(\theta+\frac{2 \pi}{3}\right)}-\frac{b y}{\sin \left(\theta+\frac{2 \pi}{3}\right)}=a^2-b^2 .....(2)\end{aligned}

\mathrm {\frac{a x}{\cos \left(\theta-\frac{2 \pi}{3}\right)}-\frac{b y}{\sin \left(\theta-\frac{2 \pi}{3}\right)}=a^2-b^2......(3)}

Normals (1), (2), (3) are concurrent if 

\mathrm {\left|\begin{array}{ccc}\sec \theta & \operatorname{cosec} \theta & 1 \\ \sec \left(\theta+\frac{2 \pi}{3}\right) & \operatorname{cosec}\left(\theta+\frac{2 \pi}{3}\right) & 1 \\ \sec \left(\theta-\frac{2 \pi}{3}\right) & \operatorname{cosec}\left(\theta-\frac{2 \pi}{3}\right) & 1\end{array}\right|=0}........(4)

On multiplying and dividing by

\begin{array}{r} \mathrm {2 \sin \theta \cos \theta \sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta-\frac{2 \pi}{3}\right) }\\\\ \mathrm {\cos \left(\theta+\frac{2 \pi}{3}\right) \cos \left(\theta-\frac{2 \pi}{3}\right)} \end{array}

Determinant in (4) can be written as

\left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin \left(\theta+\frac{2 \pi}{3}\right) & \cos \left(\theta+\frac{2 \pi}{3}\right) & \sin \left(2 \theta+\frac{4 \pi}{3}\right) \\ \sin \left(\theta-\frac{2 \pi}{3}\right) & \cos \left(\theta-\frac{2 \pi}{3}\right) & \sin \left(2 \theta-\frac{4 \pi}{3}\right) \end{array}\right|


which has zero value. Hence, normals are concurrent.

Posted by

Suraj Bhandari

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