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Let PQ be a focal chord of the parabola $y^2= 4ax$. The tangents to the parabola at P and Q meet at a point lying on the line $y = 2x + a, a > 0.$ If chord PQ subtends an angle $\theta$ at the vertex of $y^2= 4ax$, then tan$\theta= ?$

My attempt
Eqn of PQ $\rightarrow y=-2x+2a$
Solving it with parabola $$y^2−4ax(\frac{2x+y}{2a})=0$$

$$⇒y^2−4x^2−2xy=0$$

For $x^2+y^2+2hxy=0$

$$\tan\theta=|2\sqrt{h^2-ab}/a+b|$$

using this formula, I got two value of tan$\theta$ one is positive and other is negative , which is correct negative or positive value ?

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Tangents at the endpoints of a focal chord intersect on the directrix (line $x=-a$). Hence they intersect at $D=(-a,-a)$ and tangency points $P$ and $Q$ have coordinates $$ y_{P/Q}=a(-1\pm\sqrt5),\quad x_{P/Q}={a\over2}(3\mp\sqrt5). $$ Then, by the cosine rule: $$ \cos\theta={OP^2+OQ^2-PQ^2\over2\,OP\cdot OQ}=-{3\over\sqrt{29}} $$ and consequently $$ \tan\theta=-{2\sqrt{5}\over3}. $$

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  • $\begingroup$ my question is why we take the negative value of tan not positive? $\endgroup$ – Abhishek Kumar Apr 10 at 17:10
  • $\begingroup$ Because $\cos\theta<0$, while $\sin\theta>0$ as $0<\theta<180°$. $\endgroup$ – Aretino Apr 10 at 21:11
  • $\begingroup$ @AbhishekKumar If you want a comment on your solution, you should add it in detail to the question. $\endgroup$ – Aretino Apr 14 at 11:38
  • $\begingroup$ @ Aretino I have added the my solution , plz see it $\endgroup$ – Abhishek Kumar Apr 14 at 15:43
  • $\begingroup$ I see: two intersecting lines form a pair of adjacent angles, with opposite $\tan$. To get the right sign, either you draw a sketch to see which angle $POQ$ is, or you follow my method. $\endgroup$ – Aretino Apr 14 at 16:57
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Focus: $(a,0)$

$$ 2y\frac{dy}{dx}=4a\Rightarrow\frac{dy}{dx}=\pm\frac{2a}{\sqrt{4ax}}=\pm\sqrt{\frac{a}{x}} $$ and $$ \frac{dy}{dx}=\frac{2a}{y} $$ Let point $P$ have coordinates $(at^2,2at)$, then the equation of $PQ$ is $$ y=\frac{2at}{at^2-a}(x-a)=\frac{2t}{t^2-1}(x-a) $$ To find the coordinates of $Q$ we need to solve $$ \left(\frac{2t}{t^2-1}(x-a)\right)^2=4ax. $$ We do not need to solve it directly, however. Simplifying, $$ \left(\frac{2t}{t^2-1}\right)^2x^2-\left(2a\left(\frac{2t}{t^2-1}\right)^2 -4a\right)x+\left(\frac{2t}{t^2-1}\right)^2 a^2=0 $$ we know that the product of the two roots of the equation is $a^2$. So the coordinates of $Q$ can be found to be $(a/t^2,-2a/t)$.

Now note that $$ \frac{dy}{dx}=\frac{2a}{y}=\frac{2a}{2at}=1/t $$ for point $P$ and $$ \frac{dy}{dx}=-t $$ for point $Q$. When tangent at $P,Q$ intersects, $$ \frac{1}{t}(x-at^2)+2at=-t(x-\frac{a}{t^2})-\frac{2a}{t} $$ which simplifies to $x=-a$. At the point of intersection, $y=-a$, which can help us to find out that $t=\frac{1}{\sqrt2}$. (WOLOG we may ignore the negative value)

$\Rightarrow P(a/2,\sqrt2 a), Q(2a,-2\sqrt2 a)$.

$$ \tan(POQ)=\frac{\tan(POx)+\tan(QOx)}{1-\tan(POx)\tan(QOx)}=\frac{2\sqrt2+\sqrt2}{1-2\sqrt2\sqrt2}=-\sqrt2 $$ So $\theta$ is obtuse. Please tell me if there are some mistakes.

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