Question

If P and Q are the points of intersection of the straight line $x=1$ and the curve $x^{2}+xy+y^{2}=7$ then the acute angle between the normals drawn at P and Q is}

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $Tan^{-1}(\frac{15}{29})$
• D. $Tan^{-1}(\frac{5}{7})$ \bigskip

Hint:

Angle between normals is easier using slope formula rather than geometric construction.

Answer 1

The Correct Option is C

Concept: We find intersection points, compute slope of tangent, then normal direction, and finally angle between normals.

Step 1: Find intersection points.
Substitute \(x=1\) into curve: \[ 1 + y + y^{2} = 7 \] \[ y^{2} + y - 6 = 0 \] \[ (y+3)(y-2)=0 \] So: \[ P(1,2), \quad Q(1,-3) \]

Step 2: Differentiate curve implicitly.
\[ x^{2} + xy + y^{2} = 7 \] \[ 2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0 \] \[ ( x + 2y )\frac{dy}{dx} = -(2x + y) \] \[ \frac{dy}{dx} = \frac{-(2x+y)}{x+2y} \]

Step 3: Slope of normal.
Normal slope: \[ m_n = -\frac{1}{dy/dx} = \frac{x+2y}{2x+y} \]

Step 4: Compute normals at points.
At \(P(1,2)\): \[ m_1 = \frac{1+4}{2+2} = \frac{5}{4} \] At \(Q(1,-3)\): \[ m_2 = \frac{1-6}{2-3} = \frac{-5}{-1} = 5 \]

Step 5: Angle between normals.
\[ \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right| \] \[ = \left|\frac{5 - \frac{5}{4}}{1 + \frac{25}{4}}\right| = \frac{15}{29} \] \[ \theta = \tan^{-1}\left(\frac{15}{29}\right) \]