Question

The equation of the line which is a normal to the curve \(3x^2-xy+y^2=3\) at \((1,1)\) is

• A. \( 5x+y=6 \)
• B. \( x+5y=6 \)
• C. \( x-5y+4=0 \)
• D. \( 5x-y=4 \)

Hint:

Find slope of tangent (\(m_t\)) by implicit differentiation (\(dy/dx\)) and evaluating at the point.
Slope of normal (\(m_n\)) is \(-1/m_t\).
Equation of line with slope \(m_n\) through \((x_1,y_1)\) is \(y-y_1=m_n(x-x_1)\).

Answer 1

The Correct Option is B

The equation of the curve is \(3x^2-xy+y^2=3\). The point is \((1,1)\). First, check if (1,1) lies on the curve: \(3(1)^2 - (1)(1) + (1)^2 = 3 - 1 + 1 = 3\). Yes, it lies on the curve. To find the equation of the normal, we first need the slope of the tangent at (1,1). We differentiate the equation implicitly with respect to x: \( \frac{d}{dx}(3x^2-xy+y^2) = \frac{d}{dx}(3) \) \( 6x - (1 \cdot y + x \cdot \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0 \) (Using product rule for xy) \( 6x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \) \( (2y-x)\frac{dy}{dx} = y-6x \) \( \frac{dy}{dx} = \frac{y-6x}{2y-x} \) Now find the slope of the tangent at \((1,1)\) by substituting \(x=1, y=1\): Slope of tangent, \(m_t = \frac{1-6(1)}{2(1)-1} = \frac{1-6}{2-1} = \frac{-5}{1} = -5\). The normal is perpendicular to the tangent. So, the slope of the normal, \(m_n\), is: \( m_n = -\frac{1}{m_t} = -\frac{1}{-5} = \frac{1}{5} \). The equation of the normal at \((x_1, y_1)=(1,1)\) with slope \(m_n = 1/5\) is: \( y - y_1 = m_n (x - x_1) \) \( y - 1 = \frac{1}{5} (x - 1) \) \( 5(y - 1) = x - 1 \) \( 5y - 5 = x - 1 \) \( x - 5y - 1 + 5 = 0 \) \( x - 5y + 4 = 0 \) This matches option (c). Let me recheck the checkmark on the image. The image for Q54 (the one with "Wein bridge oscillator") is different. The question image I am using for Q54 is about "Normal to curve \(3x^2-xy+y^2=3\)". The checkmark seems to be on option (b) \(x+5y=6\). My derived normal is \(x-5y+4=0\). If option (b) \(x+5y=6\) is correct: Slope of this line: \(5y = -x+6 \Rightarrow y = -x/5 + 6/5\). So slope is \(-1/5\). If \(m_n = -1/5\), then \(m_t = -1/m_n = -1/(-1/5) = 5\). For \(m_t=5\), we need \( \frac{y-6x}{2y-x} = 5 \) at (1,1). \( \frac{1-6}{2-1} = \frac{-5}{1} = -5 \). So my calculated \(m_t = -5\) is correct, and \(m_n = 1/5\) is correct. The equation \(x-5y+4=0\) is correct based on this. Option (b) \(x+5y=6\). Does it pass through (1,1)? \(1+5(1)=6 \Rightarrow 1+5=6 \Rightarrow 6=6\). Yes. Its slope is \(-1/5\). This implies \(m_n = -1/5\). If \(m_n = -1/5\), then \(m_t = 5\). But we found \(m_t = -5\). So option (b) with slope \(-1/5\) is not perpendicular to a tangent with slope \(-5\). There is a discrepancy with the marked option if it's (b). My derived answer is (c). Let's assume the question text is accurate and my derivation is accurate. Then \(x-5y+4=0\) is the correct normal. This is option (c). If the intended answer was option (b) \(x+5y=6\), then slope of normal is \(-1/5\), so slope of tangent is 5. This would require \(\frac{y_1-6x_1}{2y_1-x_1} = 5\). At (1,1), this is \(\frac{1-6}{2-1} = -5 \neq 5\). So (b) is incorrect. The checkmark in the image is clearly on (b). This means there might be an error in my differentiation or the question/options/key. Recheck differentiation: \(6x - (y + x y') + 2y y' = 0 \implies 6x-y = xy' - 2yy' = y'(x-2y)\). So \(y' = \frac{6x-y}{x-2y}\). At (1,1): \(m_t = \frac{6(1)-1}{1-2(1)} = \frac{5}{-1} = -5\). (Same slope as before). Slope of normal \(m_n = 1/5\). Equation of normal: \(y-1 = \frac{1}{5}(x-1) \implies 5y-5 = x-1 \implies x-5y+4=0\). My derivation consistently leads to \(x-5y+4=0\), which is option (c). The checkmark on (b) in the provided solution image is likely incorrect. \[ \boxed{x-5y+4=0 \text{ (Matches option c; provided answer (b) seems incorrect)}} \]