The shortest distance between the curves \({\mathrm{y}}^{2}=8\mathrm{x}\) and \({x}^{2}+{y}^{2}+12y+35=0\)is : [JEE Main…

Complete the square for the circle equation:

\({x}^{2}+{y}^{2}+12y+35=0\text{ }\\ ⟹\text{  }{x}^{2}+({y}^{2}+12y)+35=0\)

Complete the square for \(y\):

\({y}^{2}+12y=(y+6{)}^{2}−36\)

Substitute back:

\({x}^{2}+(y+6{)}^{2}−36+35=0\text{ }\\ ⟹\text{  }{x}^{2}+(y+6{)}^{2}=1\)

The circle has center \((0,−6)\) and radius \(r=1\).

The parabola \({y}^{2}=8x\) can be parameterized as \((\frac{{t}^{2}}{8},t)\) (where \(t\) is a parameter).

The distance \(D\) from the circle’s center \((0,−6)\) to a point \((\frac{{t}^{2}}{8},t)\) on the parabola is:

\(D=\sqrt{{(\frac{{t}^{2}}{8}−0)}^{2}+(t−(−6){)}^{2}}\\ =\sqrt{\frac{{t}^{4}}{64}+(t+6{)}^{2}}\)

To minimize \(D\), minimize \({D}^{2}\) (since the square root is monotonic):

Take the derivative of \(f(t)\) and set it to zero:

\({f}^{′}(t)=\frac{4{t}^{3}}{64}+2(t+6)=\frac{{t}^{3}}{16}+2t+12\)

Set \({f}^{′}(t)=0\):

\(\frac{{t}^{3}}{16}+2t+12=0\text{  }⟹\text{  }{t}^{3}+32t+192=0\)

By the Rational Root Theorem, \(t=−4\) is a root (verify: \((−4{)}^{3}+32(−4)+192\)\(=−64−128+192=0\)). Factor the cubic:

\({t}^{3}+32t+192=(t+4)({t}^{2}−4t+48)\)

The quadratic \({t}^{2}−4t+48\) has no real roots (discriminant \(−176<0\)), so the only real critical point is \(t=−4\).

Evaluate \(f(t)\) at \(t=−4\):

\(f(−4)=\frac{(−4{)}^{4}}{64}+(−4+6{)}^{2}\)\(=\frac{256}{64}+{2}^{2}=4+4=8\)

Thus, \(D=\sqrt{8}=2\sqrt{2}\).

The shortest distance between the circle and the parabola is the minimum distance from the center to the parabola minus the circle’s radius:

\(\text{Shortest distance}=2\sqrt{2}−1\)