Shortest Distance btw Point & Hyperbola

[Monsu]

hi, pls anyone, how would i find the shortest distance btw a point (x,y) and a hyperbola , given the equation of the hyperbola??

 

[Tide]

Here's one way:

The slope of the line that connects the given point to the one closest on the hyperbola will be the negative reciprocal of the tangent to the point on the hyperbola (i.e. the connecting line and the tangent will be perpendicular to each other).

This should give the same result(s) as explicity writing the distance between the given point and an arbitrary point on the hyperbola them minimizing it.

 

[M98Ranger]

To find the shortest distance between a point (x,y) and a hyperbola, we can use the distance formula. The distance formula is given by d = √((x2-x1)^2 + (y2-y1)^2), where (x1,y1) and (x2,y2) are the coordinates of the two points.

First, we need to find the equation of the hyperbola. The general equation of a hyperbola is given by (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola and a and b are the distances from the center to the vertices.

We can rewrite this equation in the form of (x-h)^2/a^2 - (y-k)^2/b^2 = 1 as (x-h)^2/a^2 - (y-k)^2/b^2 = 0. Then, we can equate the distance between the point (x,y) and the hyperbola to the distance formula and solve for x and y.

We get the following equation: √((x-h)^2/a^2 - (y-k)^2/b^2) = √((x-x)^2 + (y-y)^2).

Squaring both sides and simplifying, we get (x-h)^2/a^2 - (y-k)^2/b^2 = x^2 + y^2.

Simplifying further, we get (1/a^2 - 1/b^2)x^2 + (1/a^2 - 1/b^2)y^2 = h^2/a^2 - k^2/b^2.

Now, we have an equation of a circle with center (0,0) and radius √(h^2/a^2 - k^2/b^2). The shortest distance between the point (x,y) and the hyperbola will be the shortest distance between the point (x,y) and the circle.

Using the distance formula again, we get the following equation: d = √((x-0)^2 + (y-0)^2) = √((x-x)^2 + (y-y)^2).

Simplifying, we get d = √(x^2 + y^2).

Thus, the shortest