Two straight lines one being a tangent to $y^2=4ax$ and the other to $x^2=4by$ are at right angles.Find the locus of their point of intersection.

I tried but could not reach final answer.The tangent to $y^2=4ax$ is $y=m_1x+\frac{a}{m_1}$ and the tangent to $x^2=4by$ is $x=m_2y+\frac{b}{m_2}$.Let the point of intersection of these tangents be $(h,k)$.

$k=m_1h+\frac{a}{m_1}$ and $h=m_2k+\frac{b}{m_2}$.Now i need to eliminate $m_1$ and $m_2$ from these two equations using $m_1m_2=-1$.Final answer is $(ah+bk)(h^2+k^2)+(bh-ak)^2=0$.But i could not reach final answer.

Please help me.

  • $\begingroup$ you must compute the intersection point of these two tangents as the solution of $y=m_1x+\frac{a}{m_1}$ and $x=m_2y+\frac{b}{m_2}$ $\endgroup$ – Dr. Sonnhard Graubner Aug 30 '15 at 14:24

Note that the condition we have is not $m_1m_2=-1$.

Instead, since $y=\color{red}{m_1}x+\frac{a}{m_1}$ and $x=m_2y+\frac{b}{m_2}\iff y=\color{blue}{\frac{1}{m_2}}x-\frac{b}{m_2^2}$, we have $$\color{red}{m_1}\cdot\color{blue}{\frac{1}{m_2}}=-1\iff m_2=-m_1$$

Thus, $$\begin{align}\\&k=m_1h+\frac{a}{m_1},\qquad h=m_2k+\frac{b}{m_2}=-m_1k-\frac{b}{m_1}\\&\Rightarrow m_1k=m_1^2h+a,\qquad m_1h=-m_1^2k-b\\&\Rightarrow m_1^2h-m_1k+a=0,\qquad m_1^2k+m_1h+b=0\\&\Rightarrow m_1^2hk-m_1k^2+ak=0,\qquad m_1^2kh+m_1h^2+bh=0\\&\Rightarrow (m_1^2kh=)\ m_1k^2-ak=-m_1h^2-bh\\&\Rightarrow m_1=\frac{ak-bh}{k^2+h^2}\end{align}$$

Putting this into $k=m_1h+\frac{a}{m_1}$ gives $$\begin{align}\\&k=\frac{ak-bh}{k^2+h^2}h+\frac{a(k^2+h^2)}{ak-bh}\\&\Rightarrow k(k^2+h^2)(ak-bh)=h(ak-bh)^2+a(k^2+h^2)^2\\&\Rightarrow h(ak-bh)^2-(k^2+h^2)(ak^2-bhk-ak^2-ah^2)=0\\&\Rightarrow (ak-bh)^2+(k^2+h^2)(ah+bk)=0\\&\Rightarrow (ah+bk)(h^2+k^2)+(bh-ak)^2=0\end{align}$$


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