If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x - 1, then th

Q: Wait, why do we assume the tangent at the closest point is parallel to the line?

Think of it geometrically! If the tangent were not parallel, the line would be at an angle to the curve. You could always shift the line slightly closer to the curve until it becomes tangent. At that exact moment of tangency, the line is parallel to the tangent, and the distance is minimized. It is a beautiful consequence of geometry.

Q: Can I just use the distance formula between a point $(x, x^2+4)$ and the line $4x - y - 1 = 0$?

Absolutely! You can write the distance $d$ as a function of $x$ using the perpendicular distance formula: $d = \frac{|4x - (x^2+4) - 1|}{\sqrt{4^2 + (-1)^2}}$. Then, you would minimize this function by setting its derivative to zero. However, the 'parallel tangent' method is much faster and less prone to algebraic errors. Both paths lead to the same truth!

Q: What if the line intersects the parabola? Does the 'closest point' logic still work?

Great question! If the line intersects the parabola, the shortest distance is zero. However, in JEE problems of this type, we are usually looking for the point of tangency when the line does not intersect. If the line were to intersect, the distance would be zero at the intersection points. Always check the discriminant of the quadratic equation formed by setting the line and parabola equal to see if they intersect.

Question

If $P$ is a point on the parabola $y = x^2 + 4$ which is closest to the straight line $y = 4x - 1$, then the co-ordinates of $P$ are :

Options:

(A) $(-2, 8)$

(B) $(1, 5)$

(C) $(3, 13)$

(D) $(2, 8)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a graceful, upward-opening parabola defined by the equation $y = x^2 + 4$. Below it, cutting across the plane, lies a straight line: $y = 4x - 1$.

Your goal is to find the point $P$ on the parabola that is closest to this line. This is not just an algebra problem; it is a story of optimization and geometric harmony.

### The Sliding Line Intuition

To find the shortest distance, visualize the line $y = 4x - 1$ sliding parallel to itself, moving upward toward the parabola. As it slides, it remains parallel to its original position, maintaining a constant slope of $m = 4$.

It will eventually 'kiss' the parabola at exactly one point. That point of contact is our point $P$.

Because the line is tangent to the parabola at this point, the slope of the tangent at $P$ must be identical to the slope of our line. This is the key that unlocks the entire problem.

### The Calculus Engine

Now, let us bring in the power of calculus. We have the parabola $y = x^2 + 4$.

To find the slope of the tangent at any arbitrary point $x$, we compute the derivative with respect to $x$:

$$ \frac{dy}{dx} = \frac{d}{dx}(x^2 + 4) = 2x $$

This expression, $2x$, gives us the slope of the tangent at any point on the curve. Since we know the tangent at our special point $P$ must be parallel to the line $y = 4x - 1$, we set the derivative equal to the slope of the line:

$$ 2x = 4 $$

Solving this simple linear equation, we find $x = 2$. We have found the $x$-coordinate of our point $P$.

### The Final Coordinates

With the $x$-coordinate in hand, finding the $y$-coordinate is straightforward. Since $P$ lies on the parabola, we substitute $x = 2$ back into the original equation:

$$ y = (2)^2 + 4 = 4 + 4 = 8 $$

Thus, the coordinates of the point $P$ are **$(2, 8)$**.

It is elegant, precise, and deeply satisfying. By equating the slope of the tangent to the slope of the line, we bypassed complex distance formulas and arrived directly at the solution.

Remember this: whenever you are asked for the closest point between a curve and a line, think of the parallel tangent. It is the most powerful tool in your JEE toolkit.

Animated Solution for Mathematics - Differentiation: If $P$ is a point on the parabola $y = x^{2} + 4$ which is closest to the straight line $y = 4 x - 1$ , then the co-ordinates of $P$ are :

Select Answer:

Visualized Solution

Visualizing the Curve and Line

The Shortest Distance Concept

Finding the Slope of the Line

Differentiating the Parabola

Equating the Slopes

Solving for $x$ -coordinate

Finding the $y$ -coordinate

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Sliding Line Intuition

The Calculus Engine

The Final Coordinates

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P is a point on the parabola y=x2+4 which is closest to the straight line y=4x−1, then the co-ordinates of P are :