Question

The point on the parabola ${{y}^{2}}=4x$ which are closest to the curve\[{{x}^{2}}+{{y}^{2}}-24y+128=0\] is
       $\begin{align}
  & (\text{a) }\left( 0,0 \right) \\
 & (b\text{) }\left( 2,\sqrt{2} \right) \\
 & (c\text{) }\left( 4,4 \right) \\
 & (d\text{) None} \\
\end{align}$

Answer 1

Hint: Observe the given curve \[{{x}^{2}}+{{y}^{2}}-24y+128=0\], it is the equation of a circle. Compare with the standard equation of circle and find out the centre and radius of the given circle. Next find the parametric point on the given parabola and find the equation of normal from this point on the given parabola. Then make use of the fact “the normal to the parabola should be the normal to the circle” and solve accordingly.

Complete step-by-step answer:

The curve \[{{x}^{2}}+{{y}^{2}}-24y+128=0\] is an equation of a circle.

Drawing the given parabola and the given circle, we get

The standard form of equation of a circle is,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$

Where the centre is\[~(a,b)\]and radius is \[r\].

Opening the bracket in the standard form of equation, we get

${{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{r}^{2}}=0$

On regrouping, we get

${{x}^{2}}+{{y}^{2}}-2ax-2by+\left( {{a}^{2}}+{{b}^{2}}-{{r}^{2}} \right)=0$

Comparing this equation with the given equation of the curve\[{{x}^{2}}+{{y}^{2}}-24y+128=0\], we get

$2ax=0\Rightarrow a=0$

$2by=24y$

Canceling the like terms and dividing throughout by $'2'$, we get

$\Rightarrow b=12$

And,

${{a}^{2}}+{{b}^{2}}-{{r}^{2}}=128$

Substituting the values of ‘a’ and ‘b’, we get

$0+{{12}^{2}}-{{r}^{2}}=128$

$\Rightarrow 144-128={{r}^{2}}$

$\Rightarrow {{r}^{2}}=16$

Taking square root on both sides, we get

$r=4$

So, the center and the radius of the given curve is,

$C=\left( 0,12 \right),r=4$

Now we will find the parametric points on the given parabola.

The parametric form of a standard parabola ${{y}^{2}}=4ax$ is $\left( a{{t}^{2}},2at \right)$.

Now consider the given equation of parabola,

${{y}^{2}}=4x$

Comparing this standard form of parabola, we get

$a=1$

So, the parametric form of the given parabola is $A\left( {{t}^{2}},2t \right)$.

Now we will find the normal from the point A to the given parabola.

Consider the given parabola,

${{y}^{2}}=4x$

Differentiating with respect to $'x'$, we get

$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4x \right)$

$\Rightarrow 2y\dfrac{dy}{dx}=4$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{4}{2y}=\dfrac{2}{y}$

The derivative at point $\left( {{t}^{2}},2t \right)$ is

${{\left( \dfrac{dy}{dx} \right)}_{\left( {{t}^{2}},2t \right)}}=\dfrac{2}{2t}=\dfrac{1}{t}$

Taking the reciprocal we get,

$\dfrac{dx}{dy}=t$

We know the slope of the normal is given by,

$-\dfrac{dx}{dx}$

So, the slope of the normal to the given parabola is $-t$.

Now we know the line of equation passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

Where ‘m’ is the slope of the line

So, the equation of normal to the given parabola at point $A\left( {{t}^{2}},2t \right)$and slope $m=-t$ is,

$y-2t=\left( -t \right)\left( x-{{t}^{2}} \right)\ldots \ldots .\left( i \right)$

Now to find the shortest distance between the two curves, the normal to the parabola should be the normal to the circle. And the normal to the circle will pass through the centre of the circle.

The centre of the given circle is $C\left( 0,12 \right)$. Substituting these values in the equation (i), we get

$12-2t=\left( -t \right)\left( 0-{{t}^{2}} \right)$

$\Rightarrow 12-2t={{t}^{3}}$

$\Rightarrow {{t}^{3}}+2t-12=0$

Now we will use trial and error method to find the roots of this equation.

Let\[t=0\], we get

${{0}^{3}}+2\left( 0 \right)-12$

$\Rightarrow -12\ne 0$

So, \[t=0\] is not the root of the equation.

Let \[t=1\], we get

${{1}^{3}}+2\left( 1 \right)-1$

$\Rightarrow -9\ne 0$

So, \[t=1\]is not the root of the equation.

Let ‘t=2’, we get

${{2}^{3}}+2\left( 2 \right)-12=0$

$\Rightarrow 8+4-12=0$

So, \[t=2'\]is one of the roots of the equation.

Now dividing ${{t}^{3}}+2t-12=0$ by $\left( t-2 \right)$, we get

So, ${{t}^{3}}+2t-12=0$, can be written as,

$\left( t-2 \right)\left( {{t}^{2}}+2x+6 \right)=0$

Now we will find the roots of the equation $\left( {{t}^{2}}+2x+6 \right)=0$.

Comparing this equation with the standard quadratic equation, $\left( a{{x}^{2}}+bx+c=0 \right)$, we get

$a=1,b=2,c=6$

Now we know the formula for discriminant is,

$D={{b}^{2}}-4ac$

Substituting the corresponding values, we get

$D={{2}^{2}}-4\left( 1 \right)\left( 6 \right)$

$\Rightarrow D=4-24=-20$

Now as $D<0$, so the quadratic equation will have no real roots.

So only possible real value of ‘t’ is ‘2’.

Substituting the value of ‘t’ in the parametric point \[A({{t}^{2}},2t)\] will give,

$A\left( {{2}^{2}},2\left( 2 \right) \right)=A\left( 4,4 \right)$

So, there is only one point on the parabola \[{{y}^{2}}=4x\] which is closest to the curve${{x}^{2}}+{{y}^{2}}-24y+128=0$, i.e., $\left( 4,4 \right)$.

Hence, the correct option for the given question is option (c).

Note: The shortest distance between two curves is the common normal, students often confuse and get lost.

Another approach is to find parametric points on parabola. Find a point on the given curve and find the distance between these two points. But using this method you won’t get the correct result.