If the line ax + y = c, touches both the curves x² + y² = 1 and y² = 4√2 x, then |c| is equal to : (

Q: Wait, why can't we just use the discriminant method for both?

You absolutely can! Setting the discriminant of the quadratic equation formed by substituting the line into the curve to zero is a valid, robust method. However, it often leads to messy algebra. The slope-intercept approach is more elegant and helps you build a deeper intuition for how the line interacts with the geometry of the curves.

Q: Why do we reject $t = -1$?

Remember that we substituted $t = c^2$. Since $c$ is a real number (the y-intercept of a line in the Cartesian plane), its square $c^2$ must be non-negative. Therefore, $c^2$ cannot be $-1$. This is a classic trap in JEE problems—always check the domain of your variables!

Q: Does the sign of $a$ matter in the final calculation?

Great observation! In our final equation, we deal with $a^2$. Whether $a$ is positive or negative, its square remains the same. This reflects the symmetry of the problem—there are two possible common tangents, one with a positive slope and one with a negative slope, but they both result in the same value for $|c|$.

Question

If the line $ax + y = c$, touches both the curves $x^2 + y^2 = 1$ and $y^2 = 4\sqrt{2} x$, then $|c|$ is equal to :

Options:

(A) 1/2

(B) 2

(C) $\sqrt{2}$

(D) 1/\sqrt{2}

Sigma Percentile · Accepted Answer

### The Dance of Curves: Finding the Common Tangent

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving an equation; we are witnessing a beautiful geometric dance.

We have a circle, the most symmetric of all shapes, and a parabola, the embodiment of parabolic motion. Our goal is to find a line that touches both, a common tangent. This is a classic problem that tests your ability to bridge the gap between algebraic conditions and geometric reality.

### Phase 1: The Parabola's Secret

Let us start with the parabola $y^2 = 4\sqrt{2}x$. In the world of coordinate geometry, every curve has a secret language.

For a parabola of the form $y^2 = 4Ax$, any tangent line can be expressed in the elegant slope-intercept form:

$$y = mx + \frac{A}{m}$$

By comparing our given parabola to the standard form, we identify that $4A = 4\sqrt{2}$, which means $A = \sqrt{2}$. This is our first piece of the puzzle.

Our tangent line, which we can write as $ax + y = c$ (or $y = -ax + c$), must satisfy the condition $c = \frac{A}{m}$. Since our slope $m$ is $-a$, we have the relationship $c = \frac{\sqrt{2}}{-a}$.

Squaring both sides gives us a powerful tool:

$$a^2 = \frac{2}{c^2}$$

### Phase 2: The Circle's Guard

Now, let us turn our attention to the circle $x^2 + y^2 = 1$. This circle is centered at the origin $(0,0)$ with a radius $r = 1$.

A line is tangent to a circle if and only if the perpendicular distance from the center to the line is exactly equal to the radius. The distance $d$ from the origin to the line $ax + y - c = 0$ is given by the formula:

$$d = \frac{|-c|}{\sqrt{a^2 + 1^2}}$$

Setting this equal to the radius $r = 1$, we get $\frac{|-c|}{\sqrt{a^2 + 1}} = 1$. Squaring both sides, we find the circle's condition:

$$c^2 = a^2 + 1$$

### Phase 3: The Algebraic Synthesis

We now have two conditions for the same line. From the parabola, we have $a^2 = \frac{2}{c^2}$. From the circle, we have $a^2 = c^2 - 1$.

The moment of truth arrives: we equate these two expressions for $a^2$. This gives us:

$$c^2 - 1 = \frac{2}{c^2}$$

Multiplying by $c^2$, we arrive at the polynomial $c^4 - c^2 - 2 = 0$. This is a quadratic in disguise!

Let $t = c^2$. The equation becomes $t^2 - t - 2 = 0$. Factoring this, we get $(t - 2)(t + 1) = 0$.

Since $c^2$ cannot be negative, we reject $t = -1$ and accept $t = 2$. Thus, $c^2 = 2$, which means $|c| = \sqrt{2}$.

### Conclusion

We have arrived at our destination. The absolute value of $c$ is **$\sqrt{2}$**.

This result is not just a number; it is the culmination of understanding how lines and curves interact. You have successfully navigated the geometric constraints and the algebraic manipulation. Keep this clarity of thought, and no problem will ever be too daunting.

Animated Solution for Mathematics - Conic Sections: If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 42 x$ , then $∣ c ∣$ is equal to :

Select Answer:

Visualized Solution

The Geometric Setup

Tangent to a Parabola

Applying Parabola Condition

Relating $a$ and $c$

Tangent to a Circle

Applying Circle Condition

Simplifying the Circle Condition

Combining the Conditions

Forming a Polynomial

Solving the Equation

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Dance of Curves

Phase 1

Phase 2

Phase 3

Conclusion

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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 the line ax+y=c, touches both the curves x2+y2=1 and y2=42​x, then ∣c∣ is equal to :