The equation of a common tangent to the parabolas y = x² and y = -(x - 2)² is (A) y = 4(x - 2) (B) y

Q: Wait, why did we use the slope form $y = mx - am^2$ instead of just finding the derivative $dy/dx$?

Great question! While calculus is powerful, the slope form is a 'shortcut' specifically derived for parabolas. It saves you from solving differential equations and keeps the algebra clean. In JEE, speed and accuracy are your best friends, and knowing these standard forms gives you a massive edge.

Q: Why did we get two values for $m$? Does that mean there are two common tangents?

Exactly! One value of $m$ corresponds to the x-axis ($y=0$), which is a tangent to the vertex of the first parabola. The other value gives us the specific common tangent we are looking for. Always check your roots—sometimes the math gives you more than you asked for, and you have to interpret which one fits the physical reality of the problem.

Q: What if the parabolas were shifted differently? Would the discriminant method still work?

Absolutely. The condition $D=0$ is the universal 'tangency test.' Whether the parabolas are shifted, rotated, or scaled, if a line touches a curve, the intersection must be a single point. It is the most robust tool in your geometry arsenal.

Question

The equation of a common tangent to the parabolas $y = x^2$ and $y = -(x - 2)^2$ is

Options:

(A) $y = 4(x - 2)$

(B) $y = 4(x - 1)$

(C) $y = 4(x + 1)$

(D) $y = 4(x + 2)$

Sigma Percentile · Accepted Answer

### The Dance of Two Curves

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are witnessing a geometric conversation between two parabolas.

Imagine standing on a coordinate plane. To your left, the graceful, upward-opening curve of $y = x^2$ sits at the origin. To your right, the inverted, downward-opening curve of $y = -(x - 2)^2$ waits.

Our mission is to find the bridge—the straight line that kisses both of these curves simultaneously. This is the common tangent.

### Phase 1: The Geometry of the Tangent

Before we touch a single variable, let us visualize. The first parabola, $y = x^2$, is the standard parabola with its vertex at $(0,0)$.

The second, $y = -(x - 2)^2$, is its mirror image, shifted to the right by $2$ units. A line touching both must be slanted.

It cannot be vertical, and it cannot be horizontal (except for the trivial case of the x-axis). We are looking for a line $y = mx + c$ that satisfies the tangency condition for both.

### Phase 2: The Power of the Slope Form

In JEE Advanced, time is your most precious resource. We could use calculus, setting the derivative equal to the slope $m$, but there is a more elegant path.

For any parabola of the form $x^2 = 4ay$, the equation of a tangent with slope $m$ is given by the beautiful identity $y = mx - am^2$.

Let us apply this to our first parabola, $y = x^2$. Comparing this to the standard form $x^2 = 4ay$, we see that $4a = 1$, which means $a = \frac{1}{4}$.

Substituting this into our identity, the equation of any tangent to our first parabola is simply:

$$y = mx - \frac{m^2}{4}$$

This equation is our key; it represents the family of all possible tangents to the first curve.

### Phase 3: The Bridge of Intersection

Now, here is the masterstroke. This line is not just any line; it must also be tangent to the second parabola, $y = -(x - 2)^2$.

If a line is tangent to a curve, it intersects the curve at exactly one point. Mathematically, this means that if we substitute our tangent equation into the equation of the second parabola, the resulting quadratic equation must have equal roots.

Let us perform the substitution:

$$mx - \frac{m^2}{4} = -(x - 2)^2$$

Do not be intimidated by the algebra. Expand the right side carefully: $-(x^2 - 4x + 4)$, which becomes $-x^2 + 4x - 4$.

Now, bring everything to the left side to form a standard quadratic equation in $x$:

$$x^2 + (m - 4)x + \left(4 - \frac{m^2}{4}\right) = 0$$

### Phase 4: The Discriminant Test

We are at the climax of our journey. For this quadratic to have exactly one solution, its discriminant $D$ must be zero.

Recall that for $Ax^2 + Bx + C = 0$, $D = B^2 - 4AC$. Here, $A = 1$, $B = (m - 4)$, and $C = (4 - \frac{m^2}{4})$.

Setting $D = 0$:

$$(m - 4)^2 - 4(1)\left(4 - \frac{m^2}{4}\right) = 0$$

Expanding this, we get $(m^2 - 8m + 16) - 16 + m^2 = 0$. The constants $16$ and $-16$ cancel out beautifully, leaving us with:

$$2m^2 - 8m = 0$$

Factoring this, we find $2m(m - 4) = 0$. This gives us two possibilities: $m = 0$ or $m = 4$.

### The Final Reveal

If $m = 0$, the line is $y = 0$, which is the x-axis. While technically a tangent to the vertex of the first parabola, it does not touch the second.

So, we take $m = 4$. Substituting $m = 4$ back into our tangent equation $y = mx - \frac{m^2}{4}$, we get:

$$y = 4x - \frac{16}{4}$$

This simplifies to $y = 4x - 4$. Factoring out the $4$, we arrive at the elegant solution:

**$y = 4(x - 1)$**

You have successfully navigated the geometry, the algebra, and the logic. This is the essence of JEE physics and math—taking a complex visual problem and reducing it to a single, perfect equation.

Animated Solution for Mathematics - Conic Sections: The equation of a common tangent to the parabolas $y = x^{2}$ and $y = - (x - 2)^{2}$ is

Select Answer:

Visualized Solution

Visualizing the Parabolas

General Tangent to $y = x^{2}$

Equation of Tangent in Slope Form

Condition for Common Tangency

Substituting the Line into the Parabola

Expanding the Equation

Forming the Standard Quadratic

Applying the Tangency Condition

Setting up the Discriminant

Solving for $m$

Finding the Tangent Equations

The Sigma Insight: Equation of Tangent and Normal

The Dance of Two Curves

Phase 1

Phase 2

Phase 3

Phase 4

The Final Reveal

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a common tangent to the parabolas y=x2 and y=−(x−2)2 is