Question

The normal at the point \((1,1)\) on the curve \(2y+x^2=3\) is?

• A. \(x+y=0\)
• B. \(x-y=0\)
• C. \(x+y+1=0\)
• D. \(x+y-1=0\)

Hint:

Normal slope is the negative reciprocal of tangent slope. First differentiate to get tangent slope, then find the normal equation.

Answer 1

The Correct Option is B

Concept:
The tangent slope is obtained by differentiating the curve. The normal is perpendicular to the tangent. \[ m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}} \]

Step 1: Given curve. \[ 2y+x^2=3 \]

Step 2: Differentiate with respect to \(x\). \[ \frac{d}{dx}(2y)+\frac{d}{dx}(x^2)=\frac{d}{dx}(3) \] \[ 2\frac{dy}{dx}+2x=0 \] \[ \frac{dy}{dx}=-x \]

Step 3: Find tangent slope at \((1,1)\). \[ m_{\text{tangent}}=-1 \]

Step 4: Find normal slope. \[ m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}} \] \[ m_{\text{normal}}=-\frac{1}{-1}=1 \]

Step 5: Equation of normal.
Using point-slope form: \[ y-y_1=m(x-x_1) \] At point \((1,1)\): \[ y-1=1(x-1) \] \[ y-1=x-1 \] \[ y=x \] \[ x-y=0 \] \[ \therefore \text{Correct Answer is (B)} \]