$\text{In general four normals can be drawn to the }$ $\text{rectangular hyperbola $xy = c^2$ from }$ $\text{any point P(h,k)}$
$\text{If these normals intersect the curve at }$ $A(ct_{1}, c/t_{1}), B(ct_{2}, c/t_{2}), C(ct_{3},c/t_{3}) $ $\text{and}$ $ D(ct_{4}, c/t_{4}) $respectively then A, B, C and D are known as feet of co-normal points.
The shortest distance of locus of P from the circle $x^2+y^2-24x+128=0 $ is?
My try:-
Analytically $t_{1},t_{2} ,t_{3},t_{4}$ are roots of the equation $ct^4-ht^3+ kt-c = 0.$
So taking it as f(x) we solve that f'(x)=0 must have 3 real and distinct roots and we again take a derivative f''(x) = 0 we get the roots as 0 and h/2c then $f'(0)f'(h/2c)<0$ From there I cant proceed further I am getting the locus as $y(y-x^3)=0$
But the answer is 4(√5 -1) Pls help
