Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect ea

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

You absolutely can! Using calculus is a powerful and elegant way to solve this. You could find the derivative of both curves, set the slopes equal to $m$, and solve for the point of tangency. However, the standard form $y = mx + \frac{a}{m}$ is a classic, time-saving tool in coordinate geometry that bypasses the need for differentiation, allowing you to focus on the algebraic condition of tangency directly.

Q: Why must the discriminant $D$ be zero?

Think of the intersection of a line and a parabola. Generally, a line intersects a parabola at two points. As you move the line, those two points get closer together. When the line becomes a tangent, those two points merge into one single point of contact. Algebraically, this means the quadratic equation representing the intersection must have a repeated root, which only happens when the discriminant $D = b^2 - 4ac$ is exactly zero.

Q: I got $m^3 = -1$, so $m = -1$. Are there other possible slopes?

In this specific problem, we are looking for the real slope of the common tangent. The equation $m^3 = -1$ has one real root ($m = -1$) and two complex roots. Since a geometric tangent line in the Cartesian plane must have a real slope, we reject the complex roots and proceed with $m = -1$.

Question

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

Options:

(A) $4(x + y) + 3 = 0$

(B) $8(2x + y) + 3 = 0$

(C) $3(x + y) + 4 = 0$

(D) $x + 2y + 3 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem involves two parabolas, both originating from the origin, with their axes along the coordinate axes. We are given the length of the latus rectum as $4a = 3$ and $4b = 3$, which implies $a = \frac{3}{4}$ and $b = \frac{3}{4}$.

The equations of our two parabolas are:
$$y^2 = 3x$$
$$x^2 = 3y$$

These parabolas are perfectly symmetric with respect to the line $y = x$.

### The Tangent's Identity

To find a common tangent, we consider the general equation of a tangent to the first parabola, $y^2 = 3x$. Using the standard slope-form equation $y = mx + \frac{a}{m}$ with $a = \frac{3}{4}$, we define our candidate line as:

$$y = mx + \frac{3}{4m}$$

This equation represents a family of lines, each of which is tangent to the first parabola. We must now determine which of these lines also touches the second parabola.

### The Bridge of Tangency

For this line to be a common tangent, it must also be tangent to the second parabola, $x^2 = 3y$. We substitute the expression for $y$ from our tangent equation into the equation of the second parabola:

$$x^2 = 3\left(mx + \frac{3}{4m}\right)$$

Expanding this expression, we obtain:

$$x^2 = 3mx + \frac{9}{4m}$$

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

$$x^2 - 3mx - \frac{9}{4m} = 0$$

### The Discriminant Condition

For the line to be tangent to the second parabola, the quadratic equation must have exactly one solution (a repeated root). Therefore, the discriminant $D = B^2 - 4AC$ must be equal to zero.

Substituting our coefficients $A = 1$, $B = -3m$, and $C = -\frac{9}{4m}$:

$$(-3m)^2 - 4(1)\left(-\frac{9}{4m}\right) = 0$$

$$9m^2 + \frac{9}{m} = 0$$

### Final Calculation

Solving the equation $9m^2 + \frac{9}{m} = 0$ leads to:

$$9m^2 = -\frac{9}{m}$$
$$m^3 = -1$$
$$m = -1$$

Substituting $m = -1$ back into our tangent equation $y = mx + \frac{3}{4m}$:

$$y = (-1)x + \frac{3}{4(-1)}$$
$$y = -x - \frac{3}{4}$$

Multiplying the entire equation by 4 to clear the fraction, we arrive at the final equation of the common tangent:

**$$4x + 4y + 3 = 0$$**

Select Answer:

Visualized Solution

Visualizing the Geometry

Defining the Equations

General Tangent to $y^{2} = 3 x$

Condition for Common Tangency

Forming the Quadratic Equation

Applying the Tangency Condition

Setting Discriminant to Zero

Solving for the Slope $m$

Equation of the Common Tangent

Final Simplification

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Identity

The Bridge of Tangency

The Discriminant Condition

Final Calculation

Similar Questions

The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the x-axis is

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

The ordinates of the points $P$ and $Q$ on the parabola with focus $(3, 0)$ and directrix $x = - 3$ are in the ratio $3 : 1$ . If $R (α, β)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$ , then $\frac{β ^{2}}{α}$ is equal to

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

The equations of the common tangents to the parabola $y = x^{2}$ and $y = - (x - 2)^{2}$ is/are

If $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}, m_{1} \neq = m_{2}$ are two common tangents of circle $x^{2} + y^{2} = 2$ and parabola $y^{2} = x$ , then the value of $8∣ m_{1} m_{2} ∣$ is equal to

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ is :

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