Question

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP = μ, OR=v and OQ=αv+βv, then α, β² are the roots of the equation

• A. \(3x^2-2x-1=0\)
• B. \(3x^2+2x-1=0\)
• C. \(x^2-x-2=0\)
• D. \(x^2+x-2=0\)

Answer 1

The Correct Option is C

Let \(\vec{u} = \hat{i}\) and \(\vec{OQ} = \hat{j}\). Since R is the midpoint of the arc PQ, \(\vec{OR}\) bisects the right angle \(\vec{POQ}\). We can express \(\vec{v}\) in terms of \(\vec{u}\) and \(\vec{OQ}\) (which we’ve set as \(\hat{i}\) and \(\hat{j}\) respectively).

Since R is the midpoint of the arc PQ, the vector \(\vec{OR}\) is given by:

\( \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \)

Given that \( \vec{OQ} = \alpha \vec{u} + \beta \vec{v} \), we have:

\( \hat{j} = \alpha \hat{i} + \beta \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) \)

\( \hat{j} = \left( \alpha + \frac{\beta}{\sqrt{2}} \right) \hat{i} + \frac{\beta}{\sqrt{2}} \hat{j} \)

Comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides, we get:

\( \alpha + \frac{\beta}{\sqrt{2}} = 0 \quad \text{and} \quad \frac{\beta}{\sqrt{2}} = 1 \)

From the second equation, \( \beta = \sqrt{2} \). Substituting this into the first equation gives:

\( \alpha + \frac{\sqrt{2}}{\sqrt{2}} = 0 , \quad \text{so} \quad \alpha = -1 \)

We are given that \(\alpha\) and \(\beta^2\) are the roots of a quadratic equation.

Since \(\alpha = -1\) and \(\beta = \sqrt{2}\), then \( \beta^2 = 2 \).

Let the quadratic equation be \( x^2 + bx + c = 0 \).

The sum of the roots is \( -b = \alpha + \beta^2 = -1 + 2 = 1 \), so \( b = -1 \).

The product of the roots is \( c = \alpha \cdot \beta^2 = (-1)(2) = -2 \).

Therefore, the quadratic equation is \( x^2 - x - 2 = 0 \).