The shortest distance between the curves y^2=8 x and x^2+y^2+12 y+35=0 is:

Question

The shortest distance between the curves $y^2=8 x$ and $x^2+y^2+12 y+35=0$ is:

Accepted Answer

Equation of the Normal to the Parabola:

The equation for the normal to the parabola $ y^2 = 8x $ can be expressed as:

$    y = mx - 2am - am^3 \quad 	ext{where } a = 2    $

Simplifying, we have:

$    y = mx - 4m - 2m^3    $

Center and Radius of the Circle:

The given second equation $ x^2 + y^2 + 12y + 35 = 0 $ can be rewritten to find the center and radius of the circle:

Rewrite it as $ x^2 + (y + 6)^2 = 1 $.

Thus, the center of the circle is $ C(0, -6) $ and the radius $ r = 1 $.

Finding the Slope of the Normal:

To find the point $ P $ on the parabola where the normal meets, equate:

$    -6 = -4m - 2m^3    $

Solving for $ m $ gives:

$    m = 1    $

Determine Point $ P $ on the Parabola:

Substituting $ m = 1 $ back to find $ P \left( am^2, -2am ight) $:

$    P(2, -4)    $

Calculate the Shortest Distance:

The shortest distance from point $ P(2, -4) $ to the center $ C(0, -6) $ minus the radius is:

$    CP - r = \sqrt{(2 - 0)^2 + (-4 + 6)^2} - 1 = \sqrt{4 + 4} - 1    $

$    = \sqrt{8} - 1 = 2\sqrt{2} - 1    $

Thus, the shortest distance between the given curves is $ 2\sqrt{2} - 1 $.

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