Question

If (4,4) is a point on locus whose equation is y^2=ax, prove that (16,8) is another point on the locus​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

Given (4,4) lies on $y^2=ax$

              $\Rightarrow 4^2=a(4)\Rightarrow a=\dfrac{4^2}{4}=4$

Thus the locus is $y^2=4x$

Now check the point (16,8)

          $\Rightarrow 8^2=4(16)\\\\\Rightarrow 64=64$

Hence the point (16,8) lies on the given locus