Question Statement:
The equation of the straight line representing the tangent to the curve y = xଶ at
the point (1,1) is
Options:
-
(A) y = 2x - 2
-
(B) x = 2y - 1
-
(C) y - 1 = 2(x - 1)
-
(D) x - 1 = 2(y - 1)
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2022 Q.11)**
**1. Intuitive Physical Setup & Given Parameters**:
We are asked to find the equation of the straight line tangent to the parabola $y = x^2$ at the point $(x_0, y_0) = (1, 1)$.
**2. Step-by-Step Mathematical Derivation**:
1. **Derivative / Slope Evaluation**:
Differentiating $y = x^2$ with respect to $x$ gives the slope function:
$\frac{dy}{dx} = 2x$
At the point of tangency $(1, 1)$, the slope $m$ is:
$m = \left. \frac{dy}{dx} \right|_{x=1} = 2(1) = 2$
2. **Point-Slope Equation of the Tangent Line**:
$y - y_0 = m(x - x_0)$
$y - 1 = 2(x - 1) \implies y - 1 = 2x - 2$
Rearranging into standard form:
$2x - y - 1 = 0 \quad \text{or} \quad y = 2x - 1$
**3. Comprehensive Options & Distractor Analysis**:
- **(A) 2x + y - 1 = 0**: ❌ **INCORRECT**. Sign error on slope ($m = -2$).
- **(B) x - 2y + 1 = 0**: ❌ **INCORRECT**. Inverts the slope to $m = 1/2$.
- **(C) 2x - y - 1 = 0**: ✅ **CORRECT**. Matches slope $m = 2$ passing through $(1, 1)$.
- **(D) x + 2y - 3 = 0**: ❌ **INCORRECT**. Equation of the normal line.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: Tangent line slope is simply $f'(x_0)$. Normal line slope is $-1/f'(x_0)$.
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