Question

The sum of diameters of the circles that touch (i) the parabola $75 x^2=64(5 y-3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and (ii) the y-axis, is equal to _______.

Answer 1

$x^2=\frac{64.5}{75}\left(y-\frac{3}{5}\right)$
equation of tangent at $\left(\frac{8}{5}, \frac{6}{5}\right)$
$x \cdot \frac{8}{5}=\frac{64}{15}\left(\frac{y+\frac{6}{5}}{2}-\frac{3}{5}\right)$
$3 x-4 y=0$
equation of family of circle is
$\left(x-\frac{8}{5}\right)^2+\left(y-\frac{6}{5}\right)^2+\lambda(3 x-4 y)=0$
It touches $y$ axis so $f ^2= c$
$ x^2+y^2+x\left(3 \lambda-\frac{16}{5}\right)+y\left(-4 \lambda-\frac{12}{5}\right)+4=0$
$ \frac{\left(4 \lambda+\frac{12}{5}\right)^2}{4}=4 $
$ \lambda=\frac{2}{5} \text { or } \lambda=-\frac{8}{5} $
$ \lambda=\frac{2}{5}, r=1 $
$ \lambda=-\frac{8}{5}, r=4 $
$ d_1+d_2=10$