The sum of diameters of the circles that touch (i) the parabola 75...

Given,

${x}^{2} = \dfrac{64 \times 5}{75}\left(y - \dfrac{3}{5}\right)$

Now equation of tangent at \left(\dfrac{8}{5}, \dfrac{6}{5}\right)$will be given by, </p><p>$x \times \dfrac{8}{5} = \dfrac{64}{15}\left(\dfrac{y + \dfrac{6}{5}}{2} - \dfrac{3}{5}\right)$</p><p>$⇒ 3x - 4y = 0$</p><p>Now equation of family of circle will be,</p><p>${\left(x - \dfrac{8}{5}\right)}^{2} + {\left(y - \dfrac{6}{5}\right)}^{2} + \lambda\left(3x - 4y\right) = 0$</p><p>$⇒ {x}^{3} + {y}^{3} + x\left(3\lambda - \dfrac{16}{5}\right) + y\left(- 4\lambda - \dfrac{12}{5}\right) + 4 = 0$</p><p>It touches$y$axis so$2\sqrt{{f}^{2} - c} = 0 ⇒ {f}^{2} = c$</p><p>So,$\dfrac{{\left(4\lambda + \dfrac{12}{5}\right)}^{2}}{4} = 4$</p><p>$⇒ \lambda = \dfrac{2}{5}$or$\lambda = - \dfrac{8}{5}$</p><p>If,$\lambda = \dfrac{2}{5}, \text{then}r = 1$</p><p>And if$\lambda = - \dfrac{8}{5}, \text{then}r = 4$</p><p>Now sum of diameter will be${d}_{1} + {d}_{2} = 2 \times 4 + 2 \times 1 = 10