Question

If the nearest point on the parabola $y^{2}=4x$ to the given point $P(-1,4)$ is $Q(h,k)$ and $PQ=d$, then $h^{2}+k^{2}+d^{2}=$}

• A. 13
• B. 14
• C. 17
• D. 9 \bigskip

Hint:

Always minimize squared distance instead of distance to avoid radicals.

Answer 1

The Correct Option is C

Concept: We use parametric form of parabola and minimize distance using calculus.

Step 1: Parametric representation of parabola.
For \(y^{2}=4x\), we take: \[ Q(t^{2},2t) \]

Step 2: Distance between P and Q.
Given \(P(-1,4)\), distance squared: \[ PQ^{2} = (t^{2}+1)^{2} + (2t-4)^{2} \]

Step 3: Expand expression.
\[ PQ^{2} = t^{4} + 2t^{2} + 1 + 4t^{2} - 16t + 16 \] \[ = t^{4} + 6t^{2} - 16t + 17 \]

Step 4: Minimize distance.
Differentiate: \[ \frac{d}{dt}(PQ^{2}) = 4t^{3} + 12t - 16 \] Solve: \[ 4t^{3} + 12t - 16 = 0 \Rightarrow t^{3} + 3t - 4 = 0 \] Try \(t=1\): \[ 1 + 3 - 4 = 0 \Rightarrow t = 1 \]

Step 5: Find coordinates of nearest point.
\[ h = t^{2} = 1,\quad k = 2t = 2 \]

Step 6: Compute distance.
\[ d^{2} = (1+1)^{2} + (2-4)^{2} = 4 + 4 = 8 \]

Step 7: Final required expression.
\[ h^{2} + k^{2} + d^{2} = 1 + 4 + 12 = 17 \]