Question

A circle with radius unity has its centre on the positive y-axis. If this circle touches the parabola y = 2x² tangentially at the points P and Q then the sum of the ordinates of P and Q, is

• A. $\frac{15}{4}$
• B. $\frac{15}{8}$
• C. $2\sqrt{15}$
• D. 5

Answer 1

Answer: (A)

$\frac{15}{4}$

Answer 2

To find the sum of the ordinates of the points of tangency P and Q, we will use the conditions for tangency between the circle and the parabola.

Let the equation of the circle be $x^2 + (y-k)^2 = 1^2$, where its center is $(0, k)$ on the positive y-axis and its radius is 1.
The equation of the parabola is $y = 2x^2$.

Let $(x_0, y_0)$ be a point of tangency. Due to the symmetry of the parabola about the y-axis and the center of the circle being on the y-axis, if $(x_0, y_0)$ is a point of tangency, then $(-x_0, y_0)$ will also be a point of tangency. These are the points P and Q. We need to find $y_0 + y_0 = 2y_0$.

Condition 1: The point $(x_0, y_0)$ lies on both curves.

Since $(x_0, y_0)$ lies on the parabola $y = 2x^2$: $y_0 = 2x_0^2 \quad \Rightarrow \quad x_0^2 = \frac{y_0}{2}$ (Equation 1)

Since $(x_0, y_0)$ lies on the circle $x^2 + (y-k)^2 = 1$: $x_0^2 + (y_0-k)^2 = 1$ (Equation 2)

Condition 2: The slopes of the tangents to both curves at $(x_0, y_0)$ are equal.

For the parabola $y = 2x^2$: $\frac{dy}{dx} = 4x$ The slope of the tangent to the parabola at $(x_0, y_0)$ is $m_p = 4x_0$.

For the circle $x^2 + (y-k)^2 = 1$: Differentiate implicitly with respect to x: $2x + 2(y-k)\frac{dy}{dx} = 0$ $2(y-k)\frac{dy}{dx} = -2x$ $\frac{dy}{dx} = -\frac{x}{y-k}$ The slope of the tangent to the circle at $(x_0, y_0)$ is $m_c = -\frac{x_0}{y_0-k}$.

Equating the slopes: $4x_0 = -\frac{x_0}{y_0-k}$

Since the problem states two distinct points P and Q, $x_0 \neq 0$. If $x_0=0$, then $y_0=0$, and $(0,0)$ would be the only point of tangency. We can divide by $x_0$: $4 = -\frac{1}{y_0-k}$ $4(y_0-k) = -1$ $4y_0 - 4k = -1$ $4k = 4y_0 + 1$ $k = y_0 + \frac{1}{4}$ (Equation 3)

Now, we substitute Equation 1 and Equation 3 into Equation 2: Substitute $x_0^2 = \frac{y_0}{2}$ into Equation 2: $\frac{y_0}{2} + (y_0-k)^2 = 1$

Substitute $k = y_0 + \frac{1}{4}$ into the above equation: $\frac{y_0}{2} + \left(y_0 - \left(y_0 + \frac{1}{4}\right)\right)^2 = 1$ $\frac{y_0}{2} + \left(-\frac{1}{4}\right)^2 = 1$ $\frac{y_0}{2} + \frac{1}{16} = 1$ $\frac{y_0}{2} = 1 - \frac{1}{16}$ $\frac{y_0}{2} = \frac{16-1}{16}$ $\frac{y_0}{2} = \frac{15}{16}$ $y_0 = 2 \times \frac{15}{16}$ $y_0 = \frac{15}{8}$

The ordinates of P and Q are both $y_0$. The sum of the ordinates of P and Q is $y_0 + y_0 = 2y_0$. Sum of ordinates $= 2 \times \frac{15}{8} = \frac{15}{4}$.

The final answer is $\frac{15}{4}$.