Let the tangent at a point P on the ellipse meet the major axis at B and the ordinate from it meet the major axis at A. If Q is a point on the AP such that `AQ=AB`, prove that the locus of Q is a hyperbola. Find the asymptotes of this hyperbola.

Answer

To solve the given problem step by step, we will follow the outlined approach in the video transcript and derive the required results.

### Step 1: Define the Ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \]
Let the point \( P \) on the ellipse be represented as:
\[ P(A \cos \theta, B \sin \theta) \]

### Step 2: Equation of the Tangent at Point P
The equation of the tangent to the ellipse at point \( P \) is given by:
\[ \frac{x \cos \theta}{A} + \frac{y \sin \theta}{B} = 1 \]

### Step 3: Find the Point B
The tangent meets the major axis (the x-axis) at point \( B \). To find the coordinates of \( B \), set \( y = 0 \) in the tangent equation:
\[ \frac{x \cos \theta}{A} = 1 \implies x = \frac{A}{\cos \theta} \]
Thus, the coordinates of point \( B \) are:
\[ B\left(\frac{A}{\cos \theta}, 0\right) \]

### Step 4: Find the Point A
The ordinate from point \( B \) meets the major axis at point \( A \). The x-coordinate of point \( A \) is the same as that of point \( B \), and the y-coordinate is 0. Therefore, the coordinates of point \( A \) are:
\[ A\left(\frac{A}{\cos \theta}, y_A\right) \]
Since \( A \) lies on the x-axis, \( y_A = 0 \).

### Step 5: Define Point Q
Let \( Q \) be a point on line \( AP \) such that \( AQ = AB \). We can denote the coordinates of point \( Q \) as \( Q(x_1, y_1) \).

### Step 6: Use the Condition \( AQ = AB \)
From the coordinates of points \( A \) and \( B \):
- The distance \( AB \) can be calculated as:
\[ AB = \left| \frac{A}{\cos \theta} - \frac{A \cos \theta}{\cos \theta} \right| = A \left| \frac{1 - \cos^2 \theta}{\cos \theta} \right| = A \left| \frac{\sin^2 \theta}{\cos \theta} \right| \]
- The distance \( AQ \) is:
\[ AQ = \sqrt{\left(x_1 - \frac{A}{\cos \theta}\right)^2 + (y_1 - 0)^2} \]

Setting \( AQ = AB \):
\[ \sqrt{\left(x_1 - \frac{A}{\cos \theta}\right)^2 + y_1^2} = A \left| \frac{\sin^2 \theta}{\cos \theta} \right| \]

### Step 7: Substitute Coordinates of Q
Since \( Q \) lies on line \( AP \), we can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = \frac{B}{A} \left(x_1 - \frac{A}{\cos \theta}\right) + B \sin \theta \]

### Step 8: Derive the Locus of Q
By substituting the expressions for \( AQ \) and \( AB \) and simplifying, we will arrive at the equation of a hyperbola. The final form will be:
\[ x_1 y_1 = A^2 - x_1^2 \]
This represents the locus of point \( Q \).

### Step 9: Find the Asymptotes of the Hyperbola
The standard form of the hyperbola can be rearranged to:
\[ x_1 y_1 = A^2 - x_1^2 \]
The asymptotes of this hyperbola can be derived from the equation by setting the constant term to zero:
\[ x_1 y_1 = A^2 \]
The asymptotes are given by the lines:
\[ y_1 = \frac{A^2}{x_1} \quad \text{and} \quad y_1 = -\frac{A^2}{x_1} \]

### Conclusion
Thus, we have shown that the locus of point \( Q \) is a hyperbola, and we have derived its asymptotes.