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Let $l_1$ and $l_2$ be the lengths of perpendicular chords of $y^2=4ax$ drawn through the vertex and $\left (l_1l_2 \right)^{\frac 43}= 4a^2\lambda (l_1^{\frac 23} + L_2^{\frac 23})$. Find $\lambda$

Let the chord be PQ where $P(t_1)$ and $Q(t_2)$

Since they subtend a right angle at the vertex $t_1t_2=-4$

Also Let OP be $l_1$ and OQ be $l_2$

$$PQ^2= l_1^2 +l_2^2$$ $$a^2(t_1^2-t_2^2)^2+4a^2 (t_1-t_2)^2=l_1^2+l_2^2$$ $$a^2(t_1-t_2)^2 \left [(t_1+t_2)^2+4\right ]=l_1^2+l_2^2$$

$$a^2(t_1^2+t_2^2+8)\left [t_1^2+t_2^2-4 \right ]=l_1^2+l_2^2$$

I couldnt solve further. Proceeding calculations are lengthy enough to make me think I am doing it wrong. How should I correctly solve it?

Aditya
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1 Answers1

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Take two generic perpendicular chords of $\mathcal C:y^2 = 4ax$ passing through the vertex $O=(0,0)$: \begin{gather} r:\ y=mx\\ s: \ y = -\frac{1}{m}x \end{gather} with $m> 0$. Our hope is that your condition is independent of $m$.

Find the intersection points between these lines and $\mathcal C$ in order to find $l_1$ and $l_2$. The points are: \begin{gather} P=\left(\frac{4a}{m^2},\frac{4a}{m}\right)\\ Q=\left(4am^2,-4am\right) \end{gather} And length are: \begin{gather} \overline{OP} = l_1 = \left[\left(\frac{4a}{m^2}\right)^2+\left(\frac{4a}{m}\right)^2\right]^{\frac{1}{2}} = \frac{4a}{m^2}\left(m^2+1\right)^{\frac{1}{2}}\\ \overline{OP} = l_2 = \left[\left(4am^2\right)^2+\left(4am\right)^2\right]^{\frac{1}{2}} = 4am\left(m^2+1\right)^{\frac{1}{2}} \end{gather} Now we are ready for the computation: \begin{equation} \left(l_1 l_2\right)^{\frac{4}{3}} = \left(\frac{4a}{m^2}\left(m^2+1\right)^{\frac{1}{2}}\cdot 4am\left(m^2+1\right)^{\frac{1}{2}}\right)^{\frac{4}{3}} = (4a)^{\frac{8}{3}}\left(\frac{m^2+1}{m}\right)^{\frac{4}{3}} \end{equation} And \begin{equation} 4a^2\lambda\left(l_1^{\frac{2}{3}}+ l_2^{\frac{2}{3}}\right) = 4a^2\lambda\left(4a\right)^{\frac{2}{3}}\left(m^2+1\right)^{\frac{1}{3}}\left(\frac{1}{m^{\frac{4}{3}}} + m^{\frac{2}{3}}\right) = 4^{\frac{5}{3}}(a)^{\frac{8}{3}}\lambda\left(\frac{m^2+1}{m}\right)^{\frac{4}{3}} \end{equation}

And imposing the equality, we simply obtain: \begin{gather} \lambda = 4 \end{gather}

Menezio
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