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Let $\frac{{x\cos \theta }}{2} + y\sin \theta = 1$,$\theta \in \left( {\frac{\pi }{{18}},\frac{\pi }{{15}}} \right)$ intersect the ellipse ${x^2} + 2{y^2} = 6$ at P and Q, then angle between tangents at P and Q of the ellipse is $\frac{\pi }{K}$ where K is equal to ______

My approach is as follow

$y = \frac{1}{{\sin \theta }}\left( {1 - \frac{{x\cos \theta }}{2}} \right)$

${x^2} + \frac{2}{{{{\sin }^2}\theta }}{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = 6$

${\sin ^2}\theta {x^2} + 2{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = 6{\sin ^2}\theta \Rightarrow 2{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = {\sin ^2}\theta \left( {6 - {x^2}} \right)$

$ \Rightarrow 2\left( {1 + \frac{{{x^2}{{\cos }^2}\theta }}{4} - x\cos \theta } \right) = \left( {1 - {{\cos }^2}\theta } \right)\left( {6 - {x^2}} \right) \Rightarrow \left( {2 + \frac{{{x^2}{{\cos }^2}\theta }}{2} - 2x\cos \theta } \right) = 6\left( {1 - {{\cos }^2}\theta } \right) - {x^2}\left( {1 - {{\cos }^2}\theta } \right)$

$ \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta }}{2} + \left( {1 - {{\cos }^2}\theta } \right)} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0 \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta + 2 - 2{{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0$

$\begin{array}{l} \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta }}{2} + \left( {1 - {{\cos }^2}\theta } \right)} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0 \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta + 2 - 2{{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0\\ \Rightarrow {x^2}\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 6{\cos ^2}\theta - 4 = 0 \Rightarrow {x^2} - \frac{{2x\cos \theta }}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} + \frac{{6{{\cos }^2}\theta - 4}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} = 0\end{array}$

${x_1} + {x_2} = \frac{{2\cos \theta }}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}};{x_1}{x_2} = \frac{{6{{\cos }^2}\theta - 4}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}$

${\left( {{x_1} - {x_2}} \right)^2} = \frac{{4{{\cos }^2}\theta }}{{{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}^2}}} - \frac{{\left( {24{{\cos }^2}\theta - 16} \right)}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}}$

How will I approach from here as it is getting too complicated

Samar Imam Zaidi
  • 8,800

1 Answers1

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>Alternate approach<

The equation of chord of contact of tangents drawn from an external point $(h,k)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{hx}{a^2}+\frac{ky}{b^2}=1$.

Let the external point where the tangents intersect be $P(h,k).$

$\Rightarrow$ Chord of contact of tangents from $P$ to the given ellipse: $$\frac{hx}{6}+\frac{ky}{3}=1$$

This line must coincide with the given line $\frac{x\cos\theta}2+y\sin\theta=1.$

$$\Rightarrow \frac{\frac h6}{\frac{\cos\theta}2}=\frac{\frac k3}{\sin\theta}=1$$

$$\Rightarrow P(h,k)\equiv(3\cos\theta,3\sin\theta)$$

Director circle of given ellipse: $$\Rightarrow x^2+y^2=6+3=9$$

Hence, $P$ represents any point on the director circle of the given ellipse.

$$\Rightarrow K=2$$

RiverX15
  • 888