Explanation:
B Given equation of parabola, \(\mathrm{y}^2=4 \mathrm{ax}\)
Let \((\mathrm{h}, \mathrm{k}\) ) be the midpoint.
Equation of the chord whose midpoint is \((\mathrm{h}, \mathrm{k})\) is :-
\(\mathrm{T}=\mathrm{S}_1\)
Where, \(T=y_1-2 a\left(x+x_1\right)\)
Put, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(\mathrm{h}, \mathrm{k})\)
\(\mathrm{T}=\mathrm{ky}-2 \mathrm{a}(\mathrm{x}+\mathrm{h})\)
and,
\(\mathrm{S}_1=\mathrm{y}_1{ }^2-4 \mathrm{ax}_1\)
\(\mathrm{~S}_1=\mathrm{k}^2-4 \mathrm{ah}\)
So, \(\quad[k y-2 a(x+h)]=k^2-4 a h\)
Since, It is a chord, so it must pass through focus (a, 0 ).
\(\therefore \quad {[\mathrm{k}(0)-2 \mathrm{a}(\mathrm{a}+\mathrm{h})]=\mathrm{k}^2-4 \mathrm{ah}}\)
\(\mathrm{k}^2=4 \mathrm{ah}-2 \mathrm{a}^2-2 \mathrm{ah}\)
\(\mathrm{k}^2=2 \mathrm{ah}-2 \mathrm{a}^2\)
\(\mathrm{k}^2=2 \mathrm{a}(\mathrm{h}-\mathrm{a})\)
Hence, locus is ;-
\(\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-\mathrm{a})\)