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Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.

The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$

Any point on the ellipse is given by $(a\cos \theta,b\sin \theta)$ where $a=\sqrt 3,b=\frac{\sqrt 3}{\sqrt 2}$.

Now slope of the tangent to the curve at $(a\cos \theta,b\sin \theta)$ is $\dfrac{-a\cos \theta}{2b\sin \theta}$.

Hence we have $\dfrac{b\sin \theta- 2}{a\cos \theta -1}=\dfrac{-a\cos \theta}{2b\sin \theta}$.

On simplifying we get $4b\sin \theta +a\cos\theta =3$

If we can find the value of $\theta $ from above then we can find the two points on the ellipse where the tangents touch them but I am unable to solve them.

Please help to solve it.

Any hints will be helpful

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If $y=mx+n$ is a tangent then $$n^2=a^2m^2+b^2$$ or $$n^2=3m^2+\frac{3}{2}.$$ Also, we have $2=m+n$ and we got the following equation on slopes: $$(2-m)^2=3m^2+\frac{3}{2}.$$

After this use $$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$ I got $$\alpha=\arctan12.$$

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  • $\begingroup$ Can you please explain why is $2=m+n$? $\endgroup$ – Learnmore Jan 12 '18 at 17:08
  • $\begingroup$ Because $y=mx+n$ passes through point $(1,2).$ $\endgroup$ – Michael Rozenberg Jan 12 '18 at 17:11
  • $\begingroup$ okay i get that,thank you $\endgroup$ – Learnmore Jan 12 '18 at 17:12
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 12 '18 at 17:13
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Let $m$ be the slope of a tangent to the ellipse from $(1,2)$.

The equation of the tangent is $y=mx-m+2$.

Put it into the equation of the ellipse,

\begin{align*} x^2+2(mx-m+2)^2&=3\\ (1+2m^2)x^2-4m(m-2)x+2m^2-8m+5&=0 \end{align*}

As the tangent meets the ellipse at only one point.

\begin{align*} [4m(m-2)]^2-4(1+2m^2)(2m^2-8m+5)&=0\\ (4m^4-16m^3+16m^2)-(4m^4-16m^3+12m^2-8m+5)&=0\\ 4m^2+8m-5&=0 \end{align*}

If $m_1$ and $m_2$ are the slopes of two tangents, then $\displaystyle m_1+m_2=-2$ and $\displaystyle m_1m_2=-\frac{5}{4}$.

Let $\theta$ be the acute angle between the two tangents.

\begin{align*} \tan\theta&=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\\ \tan^2\theta&=\frac{(m_1+m_2)^2-4m_1m_2}{(1+m_1m_2)^2}\\ &=\frac{(-2)^2-4(\frac{-5}{4})}{(1+\frac{-5}{4})^2}\\ &=144\\ \theta&=\tan^{-1}(12) \end{align*}

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Let $\alpha$ be the angle for which $\cos\alpha = \frac a{\sqrt{a^2+16 b^2}}$ and $\sin \alpha=\frac {4b}{\sqrt{a^2+16 b^2}}$.

Then $\theta-\alpha$ can be solved from $cos(\theta-\alpha)=\sin \alpha\sin \theta+\cos \alpha \cos \theta=\frac 3{\sqrt {a^2+16 b^2}}$

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