PQ is a double ordinate of the ellipse `x^2+9y^2 =9`, the normal at P meets the diameter through Q at R .then the locus of the mid point of PR is

Options

A. a circle
B. a parabola
C. an ellipse
D. a hyperbola

Answer

To solve the problem step by step, we will find the locus of the midpoint of segment PR, where P is a point on the ellipse, Q is a double ordinate, and R is the intersection of the normal at P with the diameter through Q.

### Step 1: Write the equation of the ellipse
The given equation of the ellipse is:
\[ x^2 + 9y^2 = 9 \]
We can rewrite it in standard form by dividing all terms by 9:
\[ \frac{x^2}{9} + \frac{y^2}{1} = 1 \]
This indicates that \( a^2 = 9 \) (so \( a = 3 \)) and \( b^2 = 1 \) (so \( b = 1 \)).

### Step 2: Parameterize the points on the ellipse
The parametric equations for the ellipse are:
\[ x = 3 \cos \theta \]
\[ y = \sin \theta \]
Thus, the coordinates of point P can be expressed as:
\[ P(3 \cos \theta, \sin \theta) \]

### Step 3: Identify the coordinates of point Q
Since PQ is a double ordinate, the x-coordinate of Q will be the same as that of P, but the y-coordinate will be \( -\sin \theta \) (as it is symmetric about the x-axis). Therefore, the coordinates of point Q are:
\[ Q(3 \cos \theta, -\sin \theta) \]

### Step 4: Find the equation of the normal at point P
The slope of the tangent line at point P can be derived from the implicit differentiation of the ellipse equation. However, for simplicity, we can directly use the normal equation:
The normal line at point P can be expressed as:
\[ y - \sin \theta = -\frac{3 \sin \theta}{3 \cos \theta} (x - 3 \cos \theta) \]
This simplifies to:
\[ y = -\tan \theta \cdot x + 3 \sin \theta + \sin \theta \]

### Step 5: Find the equation of the diameter through Q
The diameter through Q is a vertical line since Q has the same x-coordinate as P. Thus, the equation of the diameter is:
\[ x = 3 \cos \theta \]

### Step 6: Find the intersection point R
To find point R, we need to solve the equations of the normal and the diameter simultaneously. Substituting \( x = 3 \cos \theta \) into the normal equation gives:
\[ y = -\tan \theta \cdot (3 \cos \theta) + 4 \sin \theta \]
This will yield the coordinates of point R.

### Step 7: Find the midpoint M of segment PR
The midpoint M of segment PR can be calculated as:
\[ M = \left( \frac{x_P + x_R}{2}, \frac{y_P + y_R}{2} \right) \]
Substituting the coordinates of P and R into this formula will give us the coordinates of M.

### Step 8: Determine the locus of point M
To find the locus of M, we will express the coordinates of M in terms of \( \theta \) and eliminate \( \theta \). This will involve substituting the expressions for \( x_P, y_P, x_R, \) and \( y_R \) into the midpoint formula and simplifying.

### Step 9: Finalize the locus equation
After simplification, we will arrive at a relation between the coordinates of M, which will yield the equation of a conic section.