The point on curve y^2=4x which is nearest to the point (2, −8) is

The correct answer is Option (2) → (4, −4)

Let a point on \( y^{2}=4x \) be \( (t^{2},\ 2t) \). The square of its distance from (2, −8) is

    \( \displaystyle D(t)=(t^{2}-2)^{2}+(2t+8)^{2} \)

    \( \displaystyle D'(t)=2(t^{2}-2)(2t)+2(2t+8)(2) \)

    \( \displaystyle =4t^{3}-8t+8t+32 \)

    \( \displaystyle =4t^{3}+32 \)

    \( \displaystyle D'(t)=0\ \Rightarrow\ t^{3}=-8\ \Rightarrow\ t=-2 \)

    \( \displaystyle D''(t)=12t^{2}=48\gt 0\ \text{at}\ t=-2 \)

So D is least at t = −2, which gives the point \( (t^{2},2t)=(4,-4) \).

Shortcut: Test the options. (3, −3) is not on the curve, since 9 ≠ 12. For the other three, compare the squared distances from (2, −8):

    \( \displaystyle (4,-4):\ 4+16=20 \)
    \( \displaystyle (4,4):\ 4+144=148 \)
    \( \displaystyle (9,6):\ 49+196=245 \)

The least is 20, at (4, −4).