Tangents to a Family of Curves: Cracking the Code
Master systematic approaches for finding lines tangent to every curve in families like $y = x^2 + k$ with detailed JEE-level solutions.
Understanding Family of Curves Problems
In JEE, you'll encounter problems asking: "Find the equation of line(s) that are tangent to every curve in the family..." These problems test your understanding of:
- Geometric interpretation of families of curves
- Systematic elimination of the family parameter
- Condition for tangency applied to parameterized curves
- Visual reasoning about curve envelopes
🔑 Key Insight:
A line tangent to every curve in a family must be tangent to the envelope of that family. The envelope is found by eliminating the parameter between the family equation and its partial derivative.
The 3-Step Systematic Approach
Write General Tangent
Find tangent equation to a general curve in the family at point $(x_1, y_1)$
Apply Family Condition
Use the fact that $(x_1, y_1)$ lies on the family curve with parameter
Eliminate Parameter
Eliminate the family parameter to find lines valid for all curves
Problem 1: Parabolic Family
Find the equation of lines that are tangent to all parabolas of the form $y = x^2 + c$, where $c \in \mathbb{R}$.
Solution Approach:
Step 1: General tangent to $y = x^2 + c$ at point $(x_1, y_1)$:
Slope: $m = 2x_1$
Tangent equation: $y - y_1 = 2x_1(x - x_1)$
Step 2: Point $(x_1, y_1)$ lies on parabola: $y_1 = x_1^2 + c$
Substitute: $y - (x_1^2 + c) = 2x_1(x - x_1)$
Simplify: $y = 2x_1x - x_1^2 + c$
Step 3: For this to be tangent to all parabolas (all values of $c$),
The equation must be independent of $c$
This means $c$ must cancel out, which is impossible unless...
Wait! Let's think differently...
Step 4: Better approach: For a line $y = mx + k$ to be tangent to $y = x^2 + c$:
Solve: $x^2 + c = mx + k \Rightarrow x^2 - mx + (c - k) = 0$
Discriminant = 0: $m^2 - 4(c - k) = 0 \Rightarrow c = k + \frac{m^2}{4}$
But this depends on $c$! For the line to be tangent to all parabolas,
This must hold for all $c$, which is impossible.
Step 5: Conclusion: No such line exists! The family $y = x^2 + c$ are vertical translations, so no common tangent.
Problem 2: Ellipse Family
Find common tangents to all ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 + b^2 = 1$.
Solution Approach:
Step 1: General tangent to ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
$y = mx \pm \sqrt{a^2m^2 + b^2}$
Step 2: Given $a^2 + b^2 = 1$, so $b^2 = 1 - a^2$
Tangent becomes: $y = mx \pm \sqrt{a^2m^2 + 1 - a^2}$
$y = mx \pm \sqrt{a^2(m^2 - 1) + 1}$
Step 3: For this to be tangent to all ellipses in the family,
The expression under square root must be independent of $a$
$\Rightarrow m^2 - 1 = 0 \Rightarrow m = \pm 1$
Step 4: For $m = 1$: $y = x \pm \sqrt{1} = x \pm 1$
For $m = -1$: $y = -x \pm \sqrt{1} = -x \pm 1$
Step 5: Common tangents: $y = x + 1$, $y = x - 1$, $y = -x + 1$, $y = -x - 1$
Problem 3: Circle Family
Find common tangents to all circles $x^2 + y^2 - 2ax - 2ay + a^2 = 0$, where $a \in \mathbb{R}$.
Solution Approach:
Step 1: Complete the square for the circle:
$x^2 - 2ax + y^2 - 2ay + a^2 = 0$
$(x - a)^2 + (y - a)^2 = a^2$
Center: $(a, a)$, Radius: $|a|$
Step 2: A line $y = mx + c$ is tangent to a circle if distance from center to line equals radius:
$\frac{|ma + c - a|}{\sqrt{m^2 + 1}} = |a|$
Step 3: Square both sides: $\frac{(ma + c - a)^2}{m^2 + 1} = a^2$
$(ma + c - a)^2 = a^2(m^2 + 1)$
Step 4: Expand: $m^2a^2 + c^2 + a^2 + 2mca - 2mca - 2ca + \cdots$ (Wait, let's organize)
Better: Let's treat this as quadratic in $a$:
$(m-1)^2a^2 + 2c(m-1)a + c^2 = a^2(m^2 + 1)$
$[(m-1)^2 - (m^2 + 1)]a^2 + 2c(m-1)a + c^2 = 0$
$[m^2 - 2m + 1 - m^2 - 1]a^2 + 2c(m-1)a + c^2 = 0$
$-2ma^2 + 2c(m-1)a + c^2 = 0$
Step 5: For this to hold for all $a$, coefficients must be zero:
$-2m = 0 \Rightarrow m = 0$
$2c(m-1) = 0 \Rightarrow 2c(-1) = 0 \Rightarrow c = 0$
$c^2 = 0 \Rightarrow c = 0$
Step 6: Only common tangent: $y = 0$ (x-axis)
🎯 Advanced Technique: Envelope Method
For a family of curves $F(x, y, \alpha) = 0$, the envelope is found by eliminating $\alpha$ between:
$F(x, y, \alpha) = 0$ and $\frac{\partial F}{\partial \alpha} = 0$
The common tangents to the family are the tangents to this envelope!
Example: Find envelope of $y = mx + \frac{1}{m}$
Step 1: Write as $F(x, y, m) = mx - y + \frac{1}{m} = 0$
Step 2: $\frac{\partial F}{\partial m} = x - \frac{1}{m^2} = 0 \Rightarrow m^2 = \frac{1}{x}$
Step 3: Substitute back: $y = m x + \frac{1}{m} = \frac{1}{m}(m^2 x + 1)$
Step 4: Using $m^2 = \frac{1}{x}$: $y = \frac{1}{m}(\frac{x}{x} + 1) = \frac{2}{m}$
Step 5: But $m = \pm \frac{1}{\sqrt{x}}$, so envelope: $y^2 = 4x$ (a parabola!)
Problems 4-6 Available in Full Version
Includes 3 more JEE Advanced problems on tangent families with envelope method applications
📝 Quick Self-Test
Try these similar problems to test your understanding:
1. Find common tangents to all parabolas $y^2 = 4a(x - a)$
2. Find lines tangent to all curves $y = e^x + c$
3. Find common tangents to all hyperbolas $xy = c^2$
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