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 ______.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Parabola

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

75 x^{2} = 64 (5 y - 3)

at the point

(\frac{8}{5}, \frac{6}{5})

and (ii) the y-axis, is equal to ______.

Answer: 10

Rewrite the parabola:

x^{2} = \frac{64}{15} (y - \frac{3}{5})

, which is of the form

x^{2} = 4 a (y - k)

with vertex at

(0, \frac{3}{5})

and

4 a = \frac{64}{15}

.

The tangent at

(\frac{8}{5}, \frac{6}{5})

has equation:

x \cdot \frac{8}{5} = \frac{64}{15} (\frac{y + \frac{6}{5}}{2} - \frac{3}{5})

Simplifying gives

3 x - 4 y = 0

.

A circle touching the parabola at this point and the y-axis has the form:

(x - \frac{8}{5})^{2} + (y - \frac{6}{5})^{2} + λ (3 x - 4 y) = 0

Expanding:

x^{2} + y^{2} + x (3 λ - \frac{16}{5}) + y (- 4 λ - \frac{12}{5}) + 4 = 0

For tangency to the y-axis (at

x = 0

), the condition is

g^{2} = c

(\frac{3 λ - \frac{16}{5}}{2})^{2} = 4

Solving:

(4 λ + \frac{12}{5})^{2} = 16

, giving

λ = \frac{2}{5}

λ = - \frac{8}{5}

.

For

λ = \frac{2}{5}

: radius

r_{1} = 1

, diameter

= 2

.
For

λ = - \frac{8}{5}

: radius

r_{2} = 4

, diameter

= 8

.

Sum of diameters

= 2 + 8 = 10

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer