Equation of a common tangent to the parabola y² = 4x and the hyperbola xy = 2 is : (A) x + 2y + 4 =

Q: Wait, why do we use $y = mx + \frac{a}{m}$ for the parabola? Can't we just use $y = mx + c$?

You absolutely could use $y = mx + c$, but that introduces two unknowns ($m$ and $c$). By using the standard form $y = mx + \frac{a}{m}$, we immediately link the intercept $c$ to the slope $m$ using the parabola's geometry. It is a much more elegant and efficient path because it reduces the problem to solving for a single variable, $m$.

Q: Why does setting the discriminant $D=0$ work for the hyperbola?

The discriminant is the key to the nature of the roots of a quadratic equation. If $D > 0$, the line intersects the curve at two distinct points (a secant line). If $D < 0$, the line never touches the curve. When $D = 0$, the two intersection points collapse into exactly one point—which is the precise definition of a tangent. It is the algebraic way of saying the line 'just grazes' the curve.

Q: Is there only one common tangent for these two curves?

In this specific problem, our equation for $m$ resulted in $m^3 = -\frac{1}{8}$, which yields only one real root, $m = -\frac{1}{2}$. This means there is only one real common tangent. However, always be vigilant! Depending on the curves, you might find multiple real roots for $m$, which would imply multiple common tangents. Always solve the final equation for $m$ completely to ensure you haven't missed any possibilities.

Question

Equation of a common tangent to the parabola $y^2 = 4x$ and the hyperbola $xy = 2$ is :

Options:

(A) $x + 2y + 4 = 0$

(B) $x - 2y + 4 = 0$

(C) $x + y + 1 = 0$

(D) $4x + 2y + 1 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two distinct mathematical entities: the parabola $y^2 = 4x$ and the rectangular hyperbola $xy = 2$.

A common tangent is a line that performs a delicate dance, touching both curves at exactly one point each without crossing through them. It is a bridge between two different worlds of geometry.

### The Parabola's Perspective

For any parabola of the form $y^2 = 4ax$, the equation of a tangent with a given slope $m$ is elegantly defined as:

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

By comparing our parabola $y^2 = 4x$ to the standard form $y^2 = 4ax$, we immediately identify that $a = 1$.

Thus, any line that is tangent to our parabola must take the form:

$$y = mx + \frac{1}{m}$$

This expression is powerful because it encapsulates all possible tangents to the parabola in terms of a single variable, the slope $m$.

### The Hyperbola's Challenge

Now, we must force this line to also be a tangent to the hyperbola $xy = 2$. To ensure a line touches a curve, we examine their intersection.

Substituting our tangent equation $y = mx + \frac{1}{m}$ into the hyperbola equation $xy = 2$, we get:

$$x\left(mx + \frac{1}{m}\right) = 2$$

Expanding this, we arrive at:

$$mx^2 + \frac{x}{m} = 2$$

Rearranging this into the standard quadratic form $Ax^2 + Bx + C = 0$, we obtain:

$$mx^2 + \frac{1}{m}x - 2 = 0$$

### The Discriminant Bridge

This quadratic equation represents the intersection of our line and the hyperbola. For the line to be a tangent, it must touch the hyperbola at exactly one point, which requires the discriminant $D = b^2 - 4ac$ to be equal to zero.

Here, $a = m$, $b = \frac{1}{m}$, and $c = -2$. Substituting these into the discriminant formula:

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

This simplifies to:

$$\frac{1}{m^2} + 8m = 0$$

Multiplying by $m^2$ to clear the denominator, we find $1 + 8m^3 = 0$, or $m^3 = -\frac{1}{8}$. Taking the cube root, we find the slope:

$$m = -\frac{1}{2}$$

### The Final Reveal

With the slope $m = -\frac{1}{2}$ in hand, we return to our tangent equation $y = mx + \frac{1}{m}$. Substituting $m = -\frac{1}{2}$, we get:

$$y = -\frac{1}{2}x + \frac{1}{-1/2}$$

This simplifies to $y = -\frac{x}{2} - 2$. To bring this into the standard form $Ax + By + C = 0$, we multiply by $2$ to get $2y = -x - 4$.

Rearranging the terms, we arrive at the final equation of the common tangent:

**$x + 2y + 4 = 0$**

This is a beautiful result, born from the intersection of algebraic conditions and geometric intuition. You have successfully navigated the constraints of both curves to find the unique path that connects them.

Animated Solution for Mathematics - Conic Sections: Equation of a common tangent to the parabola $y^{2} = 4 x$ and the hyperbola $x y = 2$ is :

Select Answer:

Visualized Solution

Visualize the Geometry

Tangent to Parabola $y^{2} = 4 a x$

Tangent to $y^{2} = 4 x$

Condition for Common Tangency

Forming the Quadratic Equation

Applying the Tangency Condition ( $D = 0$ )

Setting up the Discriminant

Solving for the Slope $m$

Finding the Value of $m$

Final Equation of Tangent

Standard Form and Conclusion

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

Analyzing the Setup

The Parabola's Perspective

The Hyperbola's Challenge

The Discriminant Bridge

The Final Reveal

Similar Questions

The equation of the common tangent to the curves $y^{2} = 8 x$ and $x y = - 1$ is

Two common tangents to the circle $x^{2} + y^{2} = 2 a^{2}$ and parabola $y^{2} = 8 a x$ are

Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Equation of the circle with $A B$ as its diameter is

Let P be the point of intersection of the common tangents to the parabola $y^{2} = 12 x$ and the hyperbola $8 x^{2} - y^{2} = 8$ . If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

Let a line $L : 2 x + y = k, k > 0$ be a tangent to the hyperbola $x^{2} - y^{2} = 3$ . If $L$ is also a tangent to the parabola $y^{2} = α x$ , then $α$ is equal to:

From a point $A$ common tangents are drawn to the circle $x^{2} + y^{2} = a^{2} /2$ and parabola $y^{2} = 4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then

A line is a common tangent to the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ . If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2 (a + c)$ is equal to

Let $O$ be the vertex of the parabola $x^{2} = 4 y$ and $Q$ be any point on it. Let the locus of the point $P$ , which divides the line segment $O Q$ internally in the ratio $2 : 3$ be the conic $C$ . Then the equation of the chord of $C$ , which is bisected at the point $(1, 2)$ , is :

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the parabola y2=4x and the hyperbola xy=2 is :