Problem on tangent to the parabola.

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 ?

• Why you solved equation of pq with parabola( that will not surely give pair of lines intersecting at vertex) ?? Oct 31 '19 at 14:51

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 $$\tan\theta=-{2\sqrt{5}\over3}.$$

• my question is why we take the negative value of tan not positive? Apr 10 '19 at 17:10
• Because $\cos\theta<0$, while $\sin\theta>0$ as $0<\theta<180°$. Apr 10 '19 at 21:11
• @AbhishekKumar If you want a comment on your solution, you should add it in detail to the question. Apr 14 '19 at 11:38
• @ Aretino I have added the my solution , plz see it Apr 14 '19 at 15:43
• 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. Apr 14 '19 at 16:57

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.