A series of chords is drawn to the parabola $y^2=4ax$, so that their projections on a straight line which is inclined at an angle $\mathrm{\alpha }$ to the axis are all of constant length ‘$c$’ if the locus of their middle point is the curve. $\left(y^2-ax\right){\left(y\cos \alpha +2a\sin \alpha \right)}^2+a^2{c}^2=0$. Then find $\mathrm{\lambda }$

Given parabola is y² = 4ax  ....(1) 
Let M(h, k) be the midpoint of PQ.
Equation of the chord PQ passing through (h, k) and making an angle $\mathrm{\theta }$ with the x-axis is
$\frac{\mathrm{x}-\mathrm{h}}{\cos \mathrm{\theta }}=\frac{\mathrm{y}-\mathrm{k}}{\sin \mathrm{\theta }}=\mathrm{r}.$
Solving (1) & we get
${\mathrm{r}}^2{\sin }^2\mathrm{\theta }+2\mathrm{r}(\mathrm{ksin}\mathrm{\theta }-2\mathrm{aCos}\mathrm{\theta })+{\mathrm{k}}^2-4\mathrm{ah}=0$
(2) Is equal but of opposite sign as (h, k) is the middle 
(3) Point then sum of roots $=\frac{-2(\mathrm{ksin}\mathrm{\theta }-2\mathrm{acos}\mathrm{\theta })}{{\sin }^2\mathrm{\theta }}=0$
$\cot \mathrm{\theta }=\frac{\mathrm{k}}{2\mathrm{a}}\text{ than }\sin \mathrm{\theta }=\frac{2\mathrm{a}}{\sqrt{\left({\mathrm{k}}^2+4{\mathrm{a}}^2\right)}}$
Length of the chord will be 2r. 
Projection of the chord on the given line will be
$\begin{array}{l}2\mathrm{rcos}(\mathrm{\theta }-\mathrm{\alpha })=\mathrm{c}\dots \left(4\right)\text{ from }\left(3\right)\\ {\mathrm{r}}^2\frac{4{\mathrm{a}}^2}{\left({\mathrm{k}}^2+4{\mathrm{a}}^2\right)}+\left({\mathrm{k}}^2-4\mathrm{ah}\right)=0\end{array}$
Again from(4) 
$$\begin{gathered} \mathrm{r}=\frac{\mathrm{c}}{2\{\cos \mathrm{\theta \alpha }+\sin \mathrm{\theta sin}\mathrm{\alpha }\}} \\ \begin{array}{l}=\frac{\mathrm{c}}{2\sin \mathrm{\theta }\{\cos \mathrm{\alpha }\cdot \cot \mathrm{\theta }+\sin \mathrm{\alpha }\}}\\ =\frac{\mathrm{c}}{2\sin \mathrm{\theta }\left\{\cos \mathrm{\alpha }\cdot \frac{\mathrm{k}}{2\mathrm{a}}+\sin \mathrm{\alpha }\right\}}=\frac{\mathrm{c}}{\sin \mathrm{\theta }\{\mathrm{kcos}\mathrm{\alpha }+2\mathrm{asin}\mathrm{\alpha }\}}\\ \mathrm{r}=\frac{\mathrm{ac}\sqrt{\left({\mathrm{k}}^2+4{\mathrm{a}}^2\right)}}{2\mathrm{a}\{\mathrm{kcos}\mathrm{\alpha }+2\mathrm{asin}\mathrm{\theta }\{\left\}\right\}}\dots \dots .\left(6\right)\end{array} \end{gathered}$$
Form (5) &(6) we get
 $$\begin{gathered} \frac{{\mathrm{a}}^2{\mathrm{c}}^2\left({\mathrm{k}}^2+4{\mathrm{a}}^2\right)}{4{\mathrm{a}}^2[\mathrm{kcos}\mathrm{\alpha }+2\mathrm{asin}\mathrm{\alpha }]}\cdot \frac{4{\mathrm{a}}^2}{\left({\mathrm{k}}^2+4{\mathrm{a}}^2\right)}+\left({\mathrm{k}}^2-4\mathrm{ah}\right)=0 \\ \Rightarrow \left({\mathrm{k}}^2-4\mathrm{ah}\right)(\mathrm{kcos}\mathrm{\alpha }+2\mathrm{asin}\mathrm{\alpha }{)}^2+{\mathrm{a}}^2{\mathrm{c}}^2=0 \end{gathered}$$
Hence the locus of M is
$\left({\mathrm{y}}^2-4\mathrm{ax}\right)(\mathrm{ycos}\mathrm{\alpha }+2\mathrm{asin}\mathrm{\alpha }{)}^2+{\mathrm{a}}^2{\mathrm{c}}^2=0$