Question

Find the coordinates of the point on the y-axis which is nearest to the point (-2, 5).

Answer 1

Hint: Any general coordinates on y-axis can be written as (0, y), as the x-coordinate on y-axis is always zero. Use this along with the distance formula to get the nearest point to (-2, 5). After finding distance between points differentiate and equate it to 0 to get a coordinate of point which is nearest point to (-2,5).

Complete step-by-step answer:
Let A = (-2, 5) = $(x_1, y_1)$.
Now we have to find the coordinates of the point on y-axis which is nearest to the point (2, -5)
Let this point be B = $(x_2, y_2)$
Now as we know that on the y-axis the coordinate of x is zero.
Therefore $x_3$ = 0.
Therefore point B = $(0, y_2)$
Now as we know that the distance (d) between two points is calculated as
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
So the distance AB is
$AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( {0 - \left( { - 2} \right)} \right)}^2} + {{\left( {{y_2} - 5} \right)}^2}} $
Now simplify this we have,
$AB = \sqrt {4 + {{\left( {{y_2} - 5} \right)}^2}} $......................... (1)
Now according to the question we have to minimize the AB.
So differentiate equation (1) w.r.t. $y_2$ and equate to zero we have,
$ \Rightarrow \dfrac{d}{{d{y_2}}}\left( {AB} \right) = \dfrac{d}{{d{y_2}}}\sqrt {4 + {{\left( {{y_2} - 5} \right)}^2}} = 0$.................. (1)
Now as we know that differentiation of $\sqrt {a{x^2} + b} $ is
$\dfrac{d}{{dx}}\sqrt {a{x^2} + b} = \dfrac{1}{{2\sqrt {a{x^2} + b} }}\left( {\dfrac{d}{{dx}}\left( {a{x^2} + b} \right)} \right) = \dfrac{1}{{2\sqrt {ax + b} }}\left( {2ax + 0} \right) = \dfrac{{2ax}}{{2\sqrt {ax + b} }}$ so use this property in above equation we have,
Now differentiate equation (1) we have,
$ \Rightarrow \dfrac{1}{{2\sqrt {4 + {{\left( {{y_2} - 5} \right)}^2}} }}\dfrac{d}{{d{y_2}}}\left( {4 + {{\left( {{y_2} - 5} \right)}^2}} \right) = 0$
$ \Rightarrow \dfrac{1}{{\sqrt {4 + {{\left( {{y_2} - 5} \right)}^2}} }}\left( {0 + 2\left( {{y_2} - 5} \right)\left( {1 - 0} \right)} \right) = 0$
Now simplify this we have,
$ \Rightarrow 2\left( {{y_2} - 5} \right) = 0$
$ \Rightarrow {y_2} = 5$
Therefore the coordinates of B is (0, 5).
So the coordinates of the point on the y-axis which is nearest to the point (-2, 5) is (0, 5).
So this is the required answer.

Note – If we would have been told to find any general point on x-axis then eventually we could have written it (x, 0), because on x-axis the y-coordinate is zero. Students should remember the formula of distance between two points and whenever the question is asked to find closest point, we have to differentiate the distance between two points and equate it to 0.Students should be careful while differentiating the equation so we have to remember standard derivative formulas for solving these types of questions.