Question

Let the tangent and normal at the point \((3\sqrt{3},1)\) on the ellipse \(\frac{ x^2}{36}+\frac{y^2}{4} =1\) meet the y-axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line x = \(2\sqrt{5}\) intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point (α,β) , then α² - β² is equal to 

• A. 60
• B. 61
• C. \(\frac{304}{5}\)
• D. \(\frac{314}{5}\)

Answer 1

The Correct Option is C

Step 1: The ellipse and the point of tangency.
The given ellipse is \[ \frac{x^2}{36} + \frac{y^2}{4} = 1, \] and the point of tangency is \(\bigl(3\sqrt{3},\,1\bigr)\). One can verify that \(\bigl(3\sqrt{3},\,1\bigr)\) indeed lies on this ellipse. 

Step 2: Finding the tangent and normal at \(\bigl(3\sqrt{3},\,1\bigr)\).
The equation of the tangent to \(\tfrac{x^2}{36} + \tfrac{y^2}{4} = 1\) at the point \((x_1,y_1)\) is \[ \frac{x x_1}{36} + \frac{y y_1}{4} = 1. \] Hence at \(\bigl(3\sqrt{3},\,1\bigr)\), the tangent is \[ \frac{x \cdot 3\sqrt{3}}{36} + \frac{y \cdot 1}{4} = 1 \;\;\Longrightarrow\;\; \frac{x\sqrt{3}}{12} + \frac{y}{4} = 1. \] This line meets the \(y\)-axis where \(x=0\), giving \(\;\tfrac{0}{12} + \tfrac{y}{4}=1\implies y=4.\) So the tangent intersects the \(y\)-axis at \(A=(0,4)\). The slope of this tangent is found by rewriting \[ y = 4 - \frac{\sqrt{3}}{3}\,x. \] Hence the slope is \(-\tfrac{\sqrt{3}}{3}\). The normal at \(\bigl(3\sqrt{3},\,1\bigr)\) is perpendicular to this tangent, so its slope is \(\tfrac{3}{\sqrt{3}}= \sqrt{3}\). Its equation going through \(\bigl(3\sqrt{3},\,1\bigr)\) is \[ y - 1 = \sqrt{3}\,\bigl(x - 3\sqrt{3}\bigr). \] Substitute \(x=0\) to find where it meets the \(y\)-axis: \[ y - 1 = \sqrt{3}\,\bigl(0 - 3\sqrt{3}\bigr) = -9, \] so \(y=-8\). Thus \(B=(0,-8)\)

Step 3: The circle \(\mathcal{C}\) with diameter \(AB\).
The endpoints \(A=(0,4)\) and \(B=(0,-8)\). The midpoint \(O\) is \(\bigl(0,\,-2\bigr)\) and the radius is \(\tfrac{1}{2}\,AB=6\). So \(\mathcal{C}\) has center \((0,-2)\) and radius \(6\). Its equation: \[ (x-0)^2 + (y+2)^2 = 36. \] 

Step 4: Intersection of the line \(x=2\sqrt{5}\) with \(\mathcal{C}\).
Substitute \(x=2\sqrt{5}\) into the circle: \[ (2\sqrt{5})^2 + (y+2)^2 = 36 \;\;\Longrightarrow\;\; 20 + (y+2)^2 = 36 \;\;\Longrightarrow\;\; (y+2)^2 = 16 \;\;\Longrightarrow\;\; y+2 = \pm 4. \] Hence \(y=2\) or \(y=-6.\) So the points of intersection are \[ P=(2\sqrt{5},\,2),\quad Q=(2\sqrt{5},\,-6). \]

Step 5: Tangents at \(P\) and \(Q\) to the circle, and their intersection \((\alpha,\beta)\).
The tangent to the circle at a point \((x_1,y_1)\) on \((x-0)^2+(y+2)^2=36\) is: \[ x x_1 + (y+2)(y_1+2) = 36. \] At \(P=(2\sqrt{5},\,2)\), the tangent is \[ x(2\sqrt{5}) + (y+2)(2+2) = 36 \;\;\Longrightarrow\;\; 2\sqrt{5}\,x + 4(y+2) = 36. \tag{T$_P$} \] At \(Q=(2\sqrt{5},\,-6)\), the tangent is \[ x(2\sqrt{5}) + (y+2)(-6+2) = 36 \;\;\Longrightarrow\;\; 2\sqrt{5}\,x + (-4)(y+2) = 36. \tag{T$_Q$} \] We solve (T$_P$) and (T$_Q$) simultaneously: \[ \begin{aligned} &2\sqrt{5}\,x + 4(y+2) = 36,\\ &2\sqrt{5}\,x - 4(y+2) = 36. \end{aligned} \] Add these two equations: \[ 4\sqrt{5}\,x = 72 \quad\Longrightarrow\quad x = \frac{72}{4\sqrt{5}} = \frac{18}{\sqrt{5}} = \frac{18\sqrt{5}}{5}. \] Now substitute back (e.g. into T$_P$): \[ 2\sqrt{5}\cdot \frac{18\sqrt{5}}{5} + 4(y+2) = 36. \] This becomes \[ \frac{2 \cdot 18 \cdot 5}{5} + 4(y+2) = 36 \;\;\Longrightarrow\;\; 36 + 4(y+2) = 36 \;\;\Longrightarrow\;\; 4(y+2) = 0 \;\;\Longrightarrow\;\; y+2 = 0 \;\;\Longrightarrow\;\; y = -2. \] Hence the tangents intersect at \(\bigl(\tfrac{18\sqrt{5}}{5},\,-2\bigr)\). Thus \(\alpha=\tfrac{18\sqrt{5}}{5}\) and \(\beta=-2.\) 

Step 6: Calculate \(\alpha^2 - \beta^2\).
\[ \alpha^2 = \left(\frac{18\sqrt{5}}{5}\right)^2 = \frac{324 \cdot 5}{25} = \frac{1620}{25} = \frac{324}{5}, \quad \beta^2 = (-2)^2 = 4. \] Hence \[ \alpha^2 - \beta^2 = \frac{324}{5} - 4 = \frac{324}{5} - \frac{20}{5} = \frac{304}{5}. \] Hence the final result given is \[ \boxed{ \alpha^2 - \beta^2 = \frac{304}{5} }. \]