Question

The sum of coordinate of point on curve $x^2=4y$ which is at minimum distance from the line $y=x-4$ is equal to /

Answer 1

3

Answer 2

The curve is given by the equation $x^2 = 4y$. This is a parabola opening upwards with its vertex at the origin. The line is given by the equation $y = x - 4$, which can be rewritten as $x - y - 4 = 0$.

We want to find a point $(x_0, y_0)$ on the parabola $x_0^2 = 4y_0$ such that the distance from this point to the line $x - y - 4 = 0$ is minimum.

The point on the parabola closest to the line is the point where the tangent to the parabola is parallel to the given line. The slope of the given line $y = x - 4$ is $m_{line} = 1$.

Let's find the slope of the tangent to the parabola $x^2 = 4y$ at a point $(x_0, y_0)$. We can differentiate the equation of the parabola with respect to $x$: $\frac{d}{dx}(x^2) = \frac{d}{dx}(4y)$ $2x = 4 \frac{dy}{dx}$ $\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$. The slope of the tangent at the point $(x_0, y_0)$ is $\frac{x_0}{2}$.

For the tangent to be parallel to the line $y = x - 4$, their slopes must be equal: $\frac{x_0}{2} = 1$ $x_0 = 2$.

Now we need to find the corresponding $y_0$ coordinate. Since the point $(x_0, y_0)$ lies on the parabola, it must satisfy the equation $x_0^2 = 4y_0$. Substitute $x_0 = 2$ into the parabola equation: $2^2 = 4y_0$ $4 = 4y_0$ $y_0 = 1$.

So the point on the curve $x^2 = 4y$ which is at the minimum distance from the line $y = x - 4$ is $(2, 1)$.

The question asks for the sum of the coordinates of this point. Sum of coordinates = $x_0 + y_0 = 2 + 1 = 3$.

The final answer is 3.

Explanation of the solution:

1. Identify the curve $x^2=4y$ and the line $y=x-4$.
2. Find the slope of the line $y=x-4$, which is 1.
3. Find the slope of the tangent to the parabola $x^2=4y$ by differentiating the equation with respect to x: $\frac{dy}{dx} = \frac{x}{2}$.
4. The point on the parabola closest to the line has a tangent parallel to the line. Equate the slope of the tangent to the slope of the line: $\frac{x_0}{2} = 1$, which gives $x_0 = 2$.
5. Find the y-coordinate of the point by substituting $x_0=2$ into the parabola equation $x_0^2=4y_0$: $2^2 = 4y_0 \Rightarrow y_0 = 1$.
6. The point is $(2, 1)$.
7. Calculate the sum of the coordinates: $2 + 1 = 3$.

The final answer is 3.