Question

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

• A. \( 2\sqrt{3} - 1 \)
• B. \( \sqrt{2} \)
• C. \( 3\sqrt{2} - 1 \)
• D. \( 2\sqrt{2} - 1 \)

Hint:

The shortest distance between two curves often lies along the normal to one of the curves that passes through the center of the other (if it's a circle). Find the equation of the normal to the parabola, make it pass through the center of the circle, find the point of intersection, and then subtract the radius of the circle from the distance between the point and the center.

Answer 1

The Correct Option is D

Let's analyze and solve the problem of finding the shortest distance between the curves given by the equations \(y^2 = 8x\) and \(x^2 + y^2 + 12y + 35 = 0\).

The first equation represents a standard parabola \(y^2 = 8x\). This parabola opens to the right.

The second equation \(x^2 + y^2 + 12y + 35 = 0\) can be rewritten to form a circle:

• Rewrite the equation as \(x^2 + y^2 + 12y = -35\).
• Complete the square for the \(y\) terms: \(y^2 + 12y = (y + 6)^2 - 36\).
• Substitute back into the equation: \(x^2 + (y + 6)^2 - 36 = -35\).
• Simplify to get \(x^2 + (y + 6)^2 = 1\), representing a circle centered at \((0, -6)\) with a radius of \(1\).

Next, determine the shortest distance between the parabola and the circle:

• The closest point on the parabola from the x-axis is \((0, 0)\).
• Calculate the distance from the origin to the center of the circle \((0, -6)\), which is \(6\).
• Since the radius of the circle is \(1\), the shortest distance from the origin to the circle is given by the expression \(6 - 1 = 5\), as the point on the circle closest to the origin is in line with both the center and origin.

The shortest distance between the parabola \(y^2 = 8x\) and the circle \((x^2 + (y + 6)^2 = 1)\)

Hence, the correct answer is \(2\sqrt{2} - 1\).

Answer 2

The first curve is a parabola \( y^2 = 8x \). The second curve is a circle \( x^2 + y^2 + 12y + 35 = 0 \). 
Completing the square for the y terms: \( x^2 + (y^2 + 12y + 36) - 36 + 35 = 0 \) \( x^2 + (y + 6)^2 - 1 = 0 \) \( x^2 + (y + 6)^2 = 1 \) 
This is a circle with center \( C(0, -6) \) and radius \( r = 1 \). 
To find the shortest distance between the parabola and the circle, we look for a point on the parabola such that the normal at that point passes through the center of the circle. 
The equation of the parabola is \( y^2 = 8x \). 
Comparing with \( y^2 = 4ax \), we have \( 4a = 8 \Rightarrow a = 2 \). 
The equation of the normal to the parabola at the point \( (am^2, -2am) \) is \( y = mx - 2am - am^3 \). 
Substituting \( a = 2 \), the point is \( (2m^2, -4m) \) and the normal is \( y = mx - 4m - 2m^3 \). 
Since the normal passes through the center of the circle \( (0, -6) \), we substitute these coordinates into the equation of the normal: \( -6 = m(0) - 4m - 2m^3 \) \( -6 = -4m - 2m^3 \) \( 2m^3 + 4m - 6 = 0 \) \( m^3 + 2m - 3 = 0 \) 
By inspection, \( m = 1 \) is a root: \( (1)^3 + 2(1) - 3 = 1 + 2 - 3 = 0 \). So, \( (m - 1) \) is a factor. 
Dividing \( m^3 + 2m - 3 \) by \( (m - 1) \) gives \( m^2 + m + 3 \). 
The quadratic \( m^2 + m + 3 = 0 \) has discriminant \( \Delta = (1)^2 - 4(1)(3) = 1 - 12 = -11<0 \), so it has no real roots. 
Thus, the only real value of \( m \) is \( m = 1 \). 
The point P on the parabola corresponding to \( m = 1 \) is \( (2(1)^2, -4(1)) = (2, -4) \). 
The distance between the point P \( (2, -4) \) and the center of the circle C \( (0, -6) \) is: \( PC = \sqrt{(2 - 0)^2 + (-4 - (-6))^2} = \sqrt{2^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \). 
The shortest distance between the parabola and the circle is \( PC - r = 2\sqrt{2} - 1 \).