Discussion of a problem with the method of parameter isolated

Question

In the right triangle $ABC$ the hypotenuse $BC$ measures $\dfrac{5}{3}a$ and the cathetus $AC$ measures $a$. Let P be a point of the hypotenuse; It leads from $P$ to the perpendicular to the hypotenuse itself that encounters the cathetus $A$B at point $N$ and $M$ is the projection of $P$ on $AB$. Determine $PB=x$ in such a way that it is verified the report:

$\dfrac{25}{3}PM^2+9PC^2+\dfrac{16}{25}NB^2=9kAB^2$ with $k$ a parameter.

I place $x>0$ and since $PC=BC-PB=\dfrac{5}{3}a-x$, I place $x<\dfrac{5}{3}a$. The cathetus $AB=\dfrac{3}{4}a$. Then, since the triangles $ABC$ and $PBN$ are similar I find $BN=\dfrac{5}{4}x$ and in the same way $PM=\dfrac{3}{5}x$.

The report becomes $13x^2-30ax=a^2(16k-25)$. Now, I want to use the method of parameter isolated so I consider the system formed by the equations: 1) $y=13x^2-30ax$ and 2) $y=a^2(16k-25)$. The vertex of the parabola 1) is $V\bigg(\dfrac{15}{13}a; -\dfrac{225}{13}a^2\bigg)$ and since $0<x<\dfrac{5}{3}a$, I consider the arc of the parabola between points $A(0,0)$ and $B\bigg(\dfrac{5}{3}a, -\dfrac{125}{9}a^2\bigg)$. From the equation 2), for $k=\dfrac{25}{52}$, I have the tangent line to the parabola in $V$, for $k=\dfrac{25}{36}$ I have the line parallel to the axis $x$ passing for the point $B$ and for $k=\dfrac{25}{16}$ the line passing for the point $A$. I conclude by saying that there are two solutions for $25/52\leq k <25/36$ and there is on solution for $25/36<k\leq 25/16$, but the solution of the problem is: one solution for $1753/3600\leq k\leq 25/16$, where is the mistake?

Answer 1

Call $AH$ the height relative to the hypothenuse.

The problem says that the

the perpendicular to the hypotenuse itself [...] encounters the cathetus $AB$ at point $M$

thus $P$ cannot be in the projection $HC$ of the side $AC$ on $BC$. This means that $x$ is less that $BH=\dfrac{16}{15}\,a$

and this changes your discussion because for $x=0$ you get $k=\dfrac{25}{16}$ and for $x=\dfrac{16}{15}\,a$ you get $k=\dfrac{1753}{3600}$ and

for $\dfrac{16}{15}<k<\dfrac{1753}{3600}$ there is one solution, since the tangent

is when $x= \dfrac{15}{13}\,a$ that is larger than $\dfrac{16}{15}\,a$

Hope this is useful