Question

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

Answer 1

Answer: 10

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