Question

Let \( A(0, 1) \), \( B(1, 1) \), and \( C(1, 0) \) be the midpoints of the sides of a triangle with incentre at the point \( D \). If the focus of the parabola \( y^2 = 4ax \) passing through \( D \) is \( (\alpha + \beta \sqrt{3}, 0) \), where \( \alpha \) and \( \beta \) are rational numbers, then \( \frac{\alpha}{\beta^2} \) is equal to:

• A. 9/2
• B. 6
• C. 8
• D. 12

Hint:

To compute the incentre and parabola focus, carefully apply coordinate geometry formulas and simplify using rationalization and trigonometric principles.

Answer 1

The Correct Option is C

The vertices of the triangle are determined as follows:

From the midpoints \( A(0, 1) \), \( B(1, 1) \), and \( C(1, 0) \), the vertices of the triangle are:

\[ P(0, 2), \, Q(2, 2), \, R(2, 0). \]

The lengths of the sides of the triangle are:

\[ a = OP = 2, \, b = OQ = 2, \, c = PQ = 2\sqrt{2}. \]

The incentre \( D \) is given by:

\[ D = \left(\frac{aP_x + bQ_x + cR_x}{a+b+c}, \frac{aP_y + bQ_y + cR_y}{a+b+c}\right). \]

Substitute the values of \( a, b, c \):

\[ D = \left(\frac{2 \cdot 0 + 2 \cdot 2 + 2\sqrt{2} \cdot 2}{2 + 2 + 2\sqrt{2}}, \frac{2 \cdot 2 + 2 \cdot 2 + 2\sqrt{2} \cdot 0}{2 + 2 + 2\sqrt{2}}\right). \]

Simplify:

\[ D = \left(\frac{4 + 4\sqrt{2}}{2 + 2\sqrt{2}}, \frac{4}{2 + 2\sqrt{2}}\right). \]

Rationalize the denominator:

\[ D = \left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right). \]

Substitute into the parabola \( y^2 = 4ax \):

\[ \left(\frac{2}{2+\sqrt{2}}\right)^2 = 4a \cdot \frac{2}{2+\sqrt{2}}. \]

Solve for \( a \):

\[ \frac{4}{(2+\sqrt{2})^2} = 4a \cdot \frac{2}{2+\sqrt{2}}. \]

Simplify:

\[ a = \frac{1}{2(2+\sqrt{2})} = \frac{1}{4}(2-\sqrt{2}). \]

The focus of the parabola is:

\[ (\alpha + \beta \sqrt{3}, 0) = \left( \frac{2}{2+\sqrt{2}}, 0 \right). \]

Finally, calculate \( \alpha = \frac{2}{2+\sqrt{2}}, \, \beta = -\frac{1}{4} \):

\[ \frac{\alpha}{\beta^2} = \frac{\frac{2}{2+\sqrt{2}}}{\left(-\frac{1}{4}\right)^2} = 8. \]