Question

The shortest distance between the point \[\left( \dfrac{3}{2},0 \right)\] and the curve \[y=\sqrt{x},\left( x>0 \right)\] is:
\[\left( a \right)\dfrac{\sqrt{5}}{2}\]
\[\left( b \right)\dfrac{5}{4}\]
\[\left( c \right)\dfrac{3}{2}\]
\[\left( d \right)\dfrac{\sqrt{3}}{2}\]

Answer 1

Hint: We have a point \[\left( \dfrac{3}{2},0 \right)\] and we have to find the smallest distance between the curve \[y=\sqrt{x}\] and \[\left( \dfrac{3}{2},0 \right).\] So, we will assume (a, b) be the point which is at least distance as (a, b) belongs to the curve. So, we get, \[b=\sqrt{a}.\] Now, we will apply the distance formula between \[\left( \dfrac{3}{2},0 \right)\] and \[\left( a,\sqrt{a} \right)\] to find the least distance.

Complete step-by-step solution:
We are given the curve is defined as \[y=\sqrt{x}\] and the point \[\left( \dfrac{3}{2},0 \right).\] We will start by assuming that there is a point at which we get the shortest distance from the curve to the line is (a, b). Since, the point (a, b) lies on the curve \[y=\sqrt{x}\] it means that these points must satisfy the equation of the curve. So, we get, \[b=\sqrt{a}.\] So, our points are \[\left( a,b \right)=\left( a,\sqrt{a} \right).\] Now, we have that distance of \[\left( \dfrac{3}{2},0 \right)\] from \[\left( a,\sqrt{a} \right)\] is the least.
We know that distance is given by the formula, \[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.\]
Here, \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{3}{2},0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( a,\sqrt{a} \right).\] So, we get,
\[\text{Distance}=\sqrt{{{\left( a-\dfrac{3}{2} \right)}^{2}}+{{\left( \sqrt{a}-0 \right)}^{2}}}\]
Now, opening the square bracket, we get,
\[\Rightarrow \text{Distance}=\sqrt{{{a}^{2}}-3a+\dfrac{9}{4}+a}\]
Simplifying, we get,
\[\Rightarrow \text{Distance}=\sqrt{{{a}^{2}}-2a+\dfrac{9}{4}}\]
\[\Rightarrow \text{Distance}=\sqrt{{{a}^{2}}-2a+1+\dfrac{5}{4}}\]
Now, after simplification, we get,
\[\Rightarrow \text{Distance}=\sqrt{{{\left( a-1 \right)}^{2}}+\dfrac{5}{4}}\]
Also, \[{{\left( a-1 \right)}^{2}}\ge 0\]
Since distance is least which is when \[{{\left( a-1 \right)}^{2}}\] is least and the least value for \[{{\left( a-1 \right)}^{2}}\] is zero.
So, \[\text{Distance}=\sqrt{{{\left( a-1 \right)}^{2}}+\dfrac{5}{4}}\] is least when \[{{\left( a-1 \right)}^{2}}=0.\]
So, the least distance will be,
\[\Rightarrow \text{Least Distance}=\sqrt{0+\dfrac{5}{4}}\]
\[\Rightarrow \text{Least Distance}=\sqrt{\dfrac{5}{4}}\]
As, \[\sqrt{4}=2,\] we get,
\[\Rightarrow \text{Least Distance}=\dfrac{\sqrt{5}}{2}\]
Hence, the right option is (a).

Note: We can also find the point which we did at the start. Since distance is minimum when \[{{\left( a-1 \right)}^{2}}=0.\] \[{{\left( a-1 \right)}^{2}}=0\] means \[a-1=0,\] and we get a = 1. So, we get a as 1. Now, we have, \[b=\sqrt{a},\] as a = 1. So, we get, \[b=\sqrt{1}=1.\] So, we have that at (1, 1), \[y=\sqrt{x}\] has the minimum distance to the point \[\left( \dfrac{3}{2},0 \right).\] Remember that the square of any term is always positive and its lowest value is 0.