If a triangle is formed by any three tangents of the parabola $y^2=4ax$, two of whose vertices lie on the parabola $x^2=4by$, then find the locus of the third vertex.
[Math] Triangle formed by any three tangents of the parabola $y^2=4ax$
analytic geometryconic sectionsgeometrylocustriangles
Related Solutions
Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$
I think this is correct.
and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
I don't think this is correct.
Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as $$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$
So, in our case, we have $$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$
Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines, $$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$
Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.
As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.
Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).
In conclusion, the answer is $a^2\geqslant 8b^2$.
HINT.
From any point $A$ of the hyperbola, in the branch intersecting the parabola but external to it (see diagram), draw tangents $AD$ and $AF$ to the parabola, which intersect the other branch of the hyperbola at $B$ and $C$. Check that line $BC$ is also tangent to the parabola.
Best Answer
You can easily show that tangents to $y^2=4ax$ at points $t_1$ and $t_2$ intersect at the point $(at_1 t_2, a(t_1 + t_2))$. Now we are given three points $t_1, t_2, t_3$ satisfying the two relations:
$$(at_1 t_2)^2=4ab(t_1 + t_2)$$ $$(at_3 t_2)^2=4ab(t_3 + t_2)$$
And we need to find the locus of the intersection of other two tangents, that is, a relation between $x=at_1 t_3$ and $y=a(t_1 + t_3)$.
Note that, by rearranging the above equations, both $t_1$ and $t_3$ satisfy the quadratic equation $t^2(at_2^2) - 4bt - 4bt_2=0$ (assuming $t_2$ to be constant). Thus we can find $t_1 t_3$ and $t_1 + t_3$ directly from this equation which are the sum and product of roots of the quadratic. Thus:
$$t_1 + t_3 = \frac{4b}{at_2^2}$$ $$t_1t_3 = \frac{4b}{at_2}$$
Now you can substitute $x=at_1 t_3$ and $y=a(t_1 + t_3)$ in the above relations, and straightforward manipulation leaves you with the locus $x^2=4by$.