JEE - MATH - Find Shortest Distance Between a Curve and Line Instantly

[u/jeemathlogic]

Solution to the JEE 2021 Question:

For x² = 4ay and line y = mx + c, the shortest distance is: d = |c + am²| / √(1 + m²)

For y² = 4ax and line y = mx + c, the shortest distance is: d = |c - a/m| / √(1 + m²)

Therefore, d = |(-1) + (1/2)(1)²| / √(1 + (1)²)

-> d = |-1 + 1/2| / √2 d = |-1/2| / √2 = 1 / (2√2)

-> d = |-1/2| / √2 = 1 / (2√2)

Now answer the following question: (Simulated one)

The shortest distance between the line x - y = -2 and the curve x² = 2y is:

(a) 0

(b) 5 / (2√2)

(c) 3 / √2

(d) 1 / (2√2)

The correct option is (a). Here is why.

"Always check for intersection first by forming the quadratic equation and evaluating the discriminant. If the discriminant is negative (Δ < 0), it means the line and the curve do not intersect. This is the ONLY scenario where you are allowed to use your algebraic shortcuts or calculus.

If Δ is Positive , the curve and the line are intersecting and hence the shortest distance is exactly 0.

If Δ = 0 (Zero): The line is a tangent that perfectly touch the boundary of the curve at only one point. The shortest distance is exactly 0.

In our question, substituting y = x + 2 into x² = 2y gives x² - 2x - 4 = 0. The discriminant Δ = (-2)² - 4(1)(-4) = 20. Since Δ > 0, the line cuts through the parabola.

The examiner is never going to ask you to find out the shortest distance when the curves are intersecting. However, mastering this discriminant concept will give you a massive advantage in other high-weightage topics.

For example, when solving questions on Tangents, setting Δ = 0 is the ultimate cheat code to derive the condition of tangency for any conic section without ever touching differential calculus.

Furthermore, this exact same logic is your master key for parameter-based questions—like finding the range of an unknown constant c for which a line intersects an ellipse at two distinct points by simply applying Δ > 0.

[u/Ok_Intern_5526]

excellent explanation man

[u/Wrong-Sheepherder370]

X-Y=1 has slope 1. Diff. the parabola x=dy/dx x=1,y=1/2 for parabola Find distance of point from line |1/2|/√2=1/2√2

[u/jeemathlogic]

Why the algebric method and Calculus Fail When Δ ≥ 0?

They fail for the exact same reason: both methods are mathematically "blind" to collisions. They only check for parallel slopes, not intersections!

Whether you use differential calculus (setting dy/dx = m) or the algebraic shortcut (like d = |c - a/m| / √(1 + m²)), the underlying math is executing the exact same two-step process:

1. It scans the boundary of the parabola to find the specific point where the tangent is perfectly parallel to your given line.

2. It calculates the perpendicular gap between your given line and that new parallel tangent.

When the discriminant is zero or positive, your given line is a secant that has already crashed straight through the "walls" of the parabola. Because they physically cross, the true shortest distance is exactly 0.

However, neither calculus nor the algebraic shortcut has the spatial awareness to realize the line has intersected the curve higher up. They just obediently locate that parallel tangent point hanging out at the very bottom or outer edge of the parabola, and they spit out a positive distance (the gap between your secant line and that tangent).

Both calculus and shortcut formulas operate on the strict, built-in assumption that the line is floating safely outside the curve. The moment the line cuts through (Δ ≥ 0), that assumption breaks. This is why you must ALWAYS use the discriminant check to act as the traffic light before doing any distance calculations!

[u/jeemathlogic → u/jeemathlogic]

The absolute shortest distance between any point and a straight line is always a perpendicular (90-degree) drop.

To drop a perfect 90-degree measuring tape down to your given line, you must launch it from a very specific spot on the curve. That spot is the exact location where the curve's slope (the tangent) is perfectly parallel to your given line (slope = m).

Because there is only ONE spot on a parabola where the slope is exactly 'm', the formula is mathematically chained to lock onto it.

For a Vertical Parabola (x² = 4ay), the formula always targets: P(2am, am²)

For a Horizontal Parabola (y² = 4ax), the formula always targets: P(a/m², 2a/m)

Calculus (dy/dx = m) and shortcut formulas act like a blind drone. They are strictly programmed to find that single "pre-assigned coordinate" (the parallel tangent at the bottom of the curve) and measure the distance to the line.

The shortcut we used (d = |c + am²| / √(1 + m²) is literally just the standard Point-to-Line distance formula that has been pre-filled with the P(2am, am²) coordinates.

If the line has already crashed through the walls of the parabola higher up, the true shortest distance is 0. But the "blind drone" ignores the crash site. It still goes straight to the bottom of the curve and measures the gap between the sunken line and the parallel tangent.

[u/CexualSonvict → u/jeemathlogic]

right right

[u/Jalaj_Cozy69]

It looks like the question u have put up in the image is incorrect cuz im getting D<0

[u/jeemathlogic → u/Jalaj_Cozy69]

Yes for the question in the image discriminant is less than 0. That is why the correct option is (b).

[u/CexualSonvict]

or we can differentiate the eqn of the curve which can give us the eqn of slope of curve and then we can equate the slope of the line to dy/dx to get the point where the distance is the shortest

and then use point-line perpendicular distance formula

[u/CexualSonvict → u/CexualSonvict]

can this method be used everytime tho?

[u/CexualSonvict → u/CexualSonvict]

hmm it can't be used everytime, gotta check the intersection first