The line`2px+ysqrt(1-p^(2))=1(abs(p)lt1)` for different values of p, touches a fixed ellipse whose exes are the coordinate axes. Q. The locus of the point of intersection of prependicular tangents of the ellipse is

Options

A. `x^(2)+y^(2)=2`
B. `x^(2)+y^(2)=3/2`
C. `x^(2)+y^(2)=5/4`
D. `x^(2)+y^(2)=1/2`

Answer

To solve the problem step by step, we will find the locus of the point of intersection of the perpendicular tangents of the ellipse defined by the given line equation.

### Step 1: Understand the Given Line Equation
The line equation is given as:
\[ 2px + y\sqrt{1 - p^2} = 1 \quad (|p| < 1) \]
This line touches a fixed ellipse whose axes are the coordinate axes.

### Step 2: General Form of the Ellipse
The standard equation of the ellipse with axes along the coordinate axes is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 3: Rearranging the Line Equation
We can rearrange the line equation to express \(y\) in terms of \(x\):
\[ y\sqrt{1 - p^2} = 1 - 2px \]
\[ y = \frac{1 - 2px}{\sqrt{1 - p^2}} \]

### Step 4: Identify the Slope of the Tangent
The slope \(m\) of the tangent line can be identified from the rearranged equation:
\[ m = -\frac{2p}{\sqrt{1 - p^2}} \]

### Step 5: Condition for Tangency
For a line to be tangent to the ellipse, it must satisfy the condition:
\[ a^2m^2 + b^2 = 1 \]
Substituting \(m\):
\[ a^2\left(-\frac{2p}{\sqrt{1 - p^2}}\right)^2 + b^2 = 1 \]
This simplifies to:
\[ \frac{4a^2p^2}{1 - p^2} + b^2 = 1 \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 4a^2p^2 + b^2(1 - p^2) = 1 - p^2 \]
\[ 4a^2p^2 + b^2 - b^2p^2 = 1 - p^2 \]
\[ p^2(4a^2 - b^2 + 1) + b^2 - 1 = 0 \]

### Step 7: Conditions for Real \(p\)
For \(p\) to be real, the quadratic in \(p^2\) must have a discriminant of zero:
\[ 4a^2 - b^2 + 1 = 0 \]
\[ b^2 = 4a^2 + 1 \]

### Step 8: Director Circle of the Ellipse
The locus of the intersection of the perpendicular tangents to the ellipse is given by the equation of the director circle:
\[ x^2 + y^2 = a^2 + b^2 \]

### Step 9: Substitute Values of \(a^2\) and \(b^2\)
From the previous steps, we have:
\[ a^2 = \frac{1}{4}, \quad b^2 = 1 \]
Thus,
\[ x^2 + y^2 = \frac{1}{4} + 1 = \frac{5}{4} \]

### Final Answer
The locus of the point of intersection of the perpendicular tangents of the ellipse is:
\[ \boxed{x^2 + y^2 = \frac{5}{4}} \]