Question

A circle has centre 𝐶 on axes of parabola and it touches the parabola at point P. CP makes an angle of 120∘ with axis of parabola. If radius of circle is 2 , then latus rectum of parabola is

• A. 2
• B. 4
• C. 8
• D. 1

Answer 1

Answer: (A)

2

Answer 2

Let the parabola be $y^2 = 4ax$. The axis of the parabola is the x-axis. The latus rectum is $4a$. The center C of the circle lies on the axis of the parabola, so $C = (h, 0)$. Let P be the point of tangency $(x_p, y_p)$. P lies on the parabola, so $y_p^2 = 4ax_p$. The line segment CP is the radius of the circle and has length 2. So, $CP = 2$. The line CP is the normal to the parabola at P.

The slope of the normal to the parabola $y^2 = 4ax$ at a point $(x_p, y_p)$ is $m_{normal} = -\frac{y_p}{2a}$. The slope of the line segment CP connecting $C(h, 0)$ and $P(x_p, y_p)$ is $m_{CP} = \frac{y_p - 0}{x_p - h} = \frac{y_p}{x_p - h}$. Since CP is the normal, $m_{CP} = m_{normal}$: $\frac{y_p}{x_p - h} = -\frac{y_p}{2a}$ This implies $x_p - h = -2a$ (assuming $y_p \neq 0$).

We are given that CP makes an angle of 120° with the axis of the parabola (the x-axis). The slope of CP is $m_{CP} = \tan(120^\circ) = -\sqrt{3}$. $\frac{y_p}{x_p - h} = -\sqrt{3}$ Substitute $x_p - h = -2a$ into this equation: $\frac{y_p}{-2a} = -\sqrt{3} \implies y_p = 2a\sqrt{3}$ Now, substitute $y_p = 2a\sqrt{3}$ into the parabola's equation $y_p^2 = 4ax_p$: $(2a\sqrt{3})^2 = 4ax_p$ $12a^2 = 4ax_p$ Assuming $a \neq 0$ and $x_p \neq 0$, we get $x_p = 3a$.

Now we can find the coordinates of C using $x_p - h = -2a$: $3a - h = -2a \implies h = 5a$ So, the center of the circle is $C = (5a, 0)$ and the point of tangency is $P = (3a, 2a\sqrt{3})$.

The radius of the circle is given as 2. The distance CP is 2. $CP^2 = (x_p - h)^2 + (y_p - 0)^2 = (3a - 5a)^2 + (2a\sqrt{3} - 0)^2$ $CP^2 = (-2a)^2 + (2a\sqrt{3})^2 = 4a^2 + 12a^2 = 16a^2$ Since $CP = 2$, $CP^2 = 4$: $16a^2 = 4$ $a^2 = \frac{4}{16} = \frac{1}{4}$ Assuming $a > 0$, we get $a = \frac{1}{2}$. The latus rectum of the parabola $y^2 = 4ax$ is $4a$. Latus Rectum = $4a = 4 \times \frac{1}{2} = 2$.