Question

Find the shortest distance from ( 0 , 3 ) to the parabola y² = 4 x.

Answer 1

see the attachment....

I had solved it with two methods....

explanation of one method is as follows

steps to solve

1)take any point Q(x,y) on the parabola....

and using the equation of parabola....find the value of y

2)use the distance formula to take out the shortest distance in term of x

3) now differentiate it with respect to x....

4)from this we will get a critical point x

5) put this in distance formula..

6) we will get the shortest distance.....

see the attachment....

Answer 2

$\Large\boxed{\sf{\quad\sqrt2\ units\quad}}$

We're finding the shortest (perpendicular) distance of the point (0, 3) to a tangent drawn to the parabola $\sf{y^2=4x.}$

Given parabola,

$\longrightarrow\sf{y^2=4x}$

$\longrightarrow\sf{y=\pm2\sqrt x}$

According to the point (0, 3), since the shortest distance from it to the parabola has to be found, we only consider the positive root of $\sf{x}.$ The point (0, 3) is above x axis so we have to consider the part of the parabola above x axis.

$\longrightarrow\sf{y=2\sqrt x}$

The slope of the tangent to this parabola,

$\longrightarrow\sf{y'=\dfrac{d}{dx}\,\left[2\sqrt x\right]}$

$\longrightarrow\sf{y'=2\cdot\dfrac{1}{2}\cdot\dfrac{1}{\sqrt x}}$

$\longrightarrow\sf{y'=\dfrac{1}{\sqrt x}}$

For $\sf{x=h,}$

$\longrightarrow\sf{y'(h)=\dfrac{1}{\sqrt h}}$

where the shortest distance of the point (0, 3) to the parabola is assumed to be given by the distance between (0, 3) and the point $\sf{\left(h,\ 2\sqrt h\right)}$ which lies on the parabola.

Then the equation of the tangent is, by point - slope form,

$\longrightarrow\sf{y=y'\cdot x+c}$

For $\sf{x=h,}$

$\longrightarrow\sf{y(h)=y'(h)\cdot h+c}$

$\longrightarrow\sf{2\sqrt h=\dfrac{1}{\sqrt h}\cdot h+c}$

$\longrightarrow\sf{2\sqrt h=\sqrt h+c}$

$\longrightarrow\sf{c=\sqrt h}$

Hence the actual equation of the tangent is, in terms of $\sf{h,}$

$\longrightarrow\sf{y=\dfrac{x}{\sqrt h}+\sqrt h}$

$\longrightarrow\sf{\dfrac{x}{\sqrt h}-y+\sqrt h=0}$

Now the equation is in standard form, $\sf{Ax+By+C=0.}$

• $\sf{A=\dfrac{1}{\sqrt h}}$

• $\sf{B=-1}$

• $\sf{C=\sqrt h}$

We know the perpendicular distance of a point $\sf{\left(x_1,\ y_1\right)}$ to a line $\sf{Ax+By+C=0}$ is given by,

$\longrightarrow\sf{p=\dfrac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}}$

Hence the perpendicular distance of the point (0, 3) to the line $\sf{\dfrac{x}{\sqrt h}-y+\sqrt h=0}$ is,

$\longrightarrow\sf{p=\dfrac{\left|\dfrac{0}{\sqrt h}-3+\sqrt h\right|}{\sqrt{\left(\dfrac{1}{\sqrt h}\right)^2+(-1)^2}}}$

$\longrightarrow\sf{p=\left|\sqrt h-3\right|\sqrt{\dfrac{h}{h+1}}}$

But remember, it is not only perpendicular distance, but the distance between the two points (0, 3) and $\sf{\left(h,\ 2\sqrt h\right).}$ Thus,

$\longrightarrow\sf{\sqrt{(h-0)^2+\left(2\sqrt h-3\right)^2}=\left|\sqrt h-3\right|\sqrt{\dfrac{h}{h+1}}}$

$\longrightarrow\sf{h^2+\left(2\sqrt h-3\right)^2=\left(\sqrt h-3\right)^2\cdot\dfrac{h}{h+1}}$

$\longrightarrow\sf{h^2+4h-12\sqrt h+9=\left(h-6\sqrt h+9\right)\cdot\dfrac{h}{h+1}}$

$\longrightarrow\sf{\left(h^2+4h-12\sqrt h+9\right)\Big(h+1\Big)=h^2-6h\sqrt h+9h}$

$\longrightarrow\sf{h^3+4h^2-6h\sqrt h+4h-12\sqrt h+9=0}$

On taking $\sf{\sqrt h=k,}$ we get,

$\longrightarrow\sf{k^6+4k^4-6k^3+4k^2-12k+9=0}$

$\longrightarrow\sf{k^6-k^5+k^5-k^4+5k^4-5k^3-k^3+k^2+3k^2-3k-9k+9=0}$

$\longrightarrow\sf{k^5(k-1)+k^4(k-1)+5k^3(k-1)-k^2(k-1)+3k(k-1)-9(k-1)=0}$

$\longrightarrow\sf{(k-1)(k^5+k^4+5k^3-k^2+3k-9)=0}$

$\longrightarrow\sf{(k-1)(k^5-k^4+2k^4-2k^3+7k^3-7k^2+6k^2-6k+9k-9)=0}$

$\longrightarrow\sf{(k-1)(k^4(k-1)+2k^3(k-1)+7k^2(k-1)+6k(k-1)+9(k-1))=0}$

$\longrightarrow\sf{(k-1)^2(k^4+2k^3+7k^2+6k+9)=0}$

Since the equation $\sf{k^4+2k^3+7k^2+6k+9=0}$ has no real roots, we consider,

$\longrightarrow\sf{(k-1)^2=0}$

$\longrightarrow\sf{k=1}$

$\longrightarrow\sf{\sqrt h=1}$

Hence the shortest distance of the point (0, 3) to the parabola is,

$\longrightarrow\sf{p=\left|1-3\right|\sqrt{\dfrac{1^2}{1^2+1}}}$

$\longrightarrow\sf{p=2\cdot\sqrt{\dfrac{1}{2}}}$

$\longrightarrow\sf{\underline{\underline{p=\sqrt2\ units}}}$