If y = mx + 4 is a tangent to both the parabolas y² = 4x and x² = 2by. Then value of b is (A) -64 (B

Q: Wait, why can't we just use the condition $c = \frac{a}{m}$ for the second parabola too?

That is a fantastic question! The condition $c = \frac{a}{m}$ is specifically derived for the standard parabola $y^2 = 4ax$. Your second parabola is $x^2 = 2by$, which is of the form $x^2 = 4Ay$. The condition of tangency for this form is actually $c = -Am^2$. While you could memorize this, using the discriminant method ($D=0$) is a universal tool that works for any conic section, making it much safer and more reliable in the exam hall.

Q: Why do we reject $b = 0$ at the end? It is a valid solution to the quadratic equation.

Algebraically, $b = 0$ is a solution to the equation $b^2 + 128b = 0$. However, we must always check the physical validity of our solutions in geometry. If you substitute $b = 0$ into the equation $x^2 = 2by$, you get $x^2 = 0$, which simplifies to $x = 0$. This is the equation of the y-axis—a straight line, not a parabola. Since the problem explicitly defines the curve as a parabola, $b = 0$ is a degenerate case that we must discard.

Q: Is there any way to visualize why the discriminant must be zero?

Absolutely! Think of the discriminant as a 'collision detector.' When you substitute the line into the parabola, you are looking for intersection points. If $D > 0$, the line cuts through the parabola at two distinct points (a secant). If $D < 0$, the line misses the parabola entirely. When $D = 0$, the two intersection points 'collide' and merge into one single point of contact. That is the very definition of a tangent!

Question

If $y = mx + 4$ is a tangent to both the parabolas $y^2 = 4x$ and $x^2 = 2by$. Then value of $b$ is

Options:

(A) -64

(B) -32

(C) -128

(D) 16

Sigma Percentile · Accepted Answer

### The Geometry of Connection: Unveiling the Common Tangent

Welcome, future engineer. Today, we are not just solving an algebra problem; we are exploring the elegant architecture of conic sections.

We are tasked with finding a common tangent—a line that serves as a bridge between two different worlds: the right-opening parabola $y^2 = 4x$ and the vertically-oriented parabola $x^2 = 2by$. This is a classic JEE Advanced problem that tests your ability to bridge algebraic conditions with geometric intuition.

### Phase 1: The First Anchor

Let us start with the first parabola, $y^2 = 4x$. In the language of coordinate geometry, this is the standard form $y^2 = 4ax$. By simple inspection, we identify that $4a = 4$, which gives us $a = 1$.

We are given a line $y = mx + 4$ that is tangent to this curve. Now, recall the fundamental condition of tangency for a parabola of the form $y^2 = 4ax$. A line $y = mx + c$ touches this parabola if and only if $c = \frac{a}{m}$.

This is not just a formula; it is a geometric constraint that ensures the line kisses the curve at exactly one point. We know our y-intercept $c$ is $4$, and our parameter $a$ is $1$. Substituting these into our condition, we get $4 = \frac{1}{m}$.

Solving for $m$, we find the slope of our bridge: $m = \frac{1}{4}$. Our tangent line is now fully defined: $y = \frac{1}{4}x + 4$.

### Phase 2: The Collision Course

Now, we turn our attention to the second parabola, $x^2 = 2by$. This curve is the mystery. We know our line $y = \frac{1}{4}x + 4$ must also be tangent to this curve.

To find the point of contact, we must force the line and the parabola to intersect. We do this by substituting the expression for $y$ from our line into the equation of the parabola:

$$x^2 = 2b \left( \frac{1}{4}x + 4 \right)$$

Let us expand this carefully. Multiplying $2b$ into the parenthesis, we get:

$$x^2 = \frac{b}{2}x + 8b$$

To analyze this intersection, we bring all terms to one side to form a standard quadratic equation in $x$:

$$x^2 - \frac{b}{2}x - 8b = 0$$

### Phase 3: The Discriminant as a Gatekeeper

Here is where the magic happens. A line is tangent to a curve if it intersects the curve at exactly one point. In the language of quadratic equations, this means our equation $x^2 - \frac{b}{2}x - 8b = 0$ must have equal roots.

For any quadratic equation $Ax^2 + Bx + C = 0$, the condition for equal roots is that the discriminant $D = B^2 - 4AC$ must be exactly zero. Let us identify our coefficients: $A = 1$, $B = -\frac{b}{2}$, and $C = -8b$.

Plugging these into our discriminant formula:

$$D = \left( -\frac{b}{2} \right)^2 - 4(1)(-8b) = 0$$

Squaring the first term gives us $\frac{b^2}{4}$, and the second term becomes $+32b$. So, we have:

$$\frac{b^2}{4} + 32b = 0$$

To make our lives easier and eliminate the fraction, we multiply the entire equation by $4$:

$$b^2 + 128b = 0$$

### Phase 4: The Final Verdict

We are almost there. Factoring out $b$, we get:

$$b(b + 128) = 0$$

This gives us two potential candidates for $b$: $b = 0$ or $b = -128$. But wait! We must pause and verify.

If $b = 0$, the equation $x^2 = 2by$ collapses into $x^2 = 0$, which is just the y-axis. That is not a parabola. Therefore, we must reject $b = 0$.

The only valid, physically meaningful solution is **$b = -128$**.

Animated Solution for Mathematics - Conic Sections: If $y = m x + 4$ is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ . Then value of $b$ is

Select Answer:

Visualized Solution

Identify the First Parabola

Condition of Tangency

Solve for Slope $m$

Consider the Second Parabola

Substitution Method

Simplify the Equation

The Tangency Condition ( $D = 0$ )

Calculate the Discriminant

Eliminate the Fraction

Factorize to Find $b$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Connection

Phase 1

Phase 2

Phase 3

Phase 4

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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; } If y=mx+4 is a tangent to both the parabolas y2=4x and x2=2by. Then value of b is