Given:
The table of value and y is inversely proportional to square of x.
To find:
The value of c and d.
Solution:
It is given that y is inversely proportional to square of x. So,
[tex]y\propto \dfrac{1}{x^2}[/tex]
[tex]y=k\dfrac{1}{x^2}[/tex] ...(i)
Where, k is the contant of proportionality.
From the given table it is clear that [tex]y=4[/tex] and [tex]x=3[/tex]. Putting these values in (i), we get
[tex]4=k\dfrac{1}{3^2}[/tex]
[tex]4=\dfrac{k}{9}[/tex]
[tex]4\times 9=k[/tex]
[tex]36=k[/tex]
Putting [tex]k=36[/tex] in (i), we get
[tex]y=36\dfrac{1}{x^2}[/tex]
[tex]y=\dfrac{36}{x^2}[/tex] ...(ii)
Putting [tex]x=5,y=c[/tex] in (ii), we get
[tex]c=\dfrac{36}{(5)^2}[/tex]
[tex]c=\dfrac{36}{25}[/tex]
Putting [tex]x=d,y=2[/tex] in (ii), we get
[tex]2=\dfrac{36}{d^2}[/tex]
[tex]d^2=\dfrac{36}{2}[/tex]
[tex]d^2=18[/tex]
Taking square root on both sides, we get
[tex]d=\pm \sqrt{18}[/tex]
[tex]d=\pm 3\sqrt{2}[/tex]
Therefore, the required values are [tex]c=\dfrac{36}{25}[/tex] and [tex]d=\pm 3\sqrt{2}[/tex].
