If the perpendicular drawn from P on the polar of P with respect to the parabola `y^2 = 4ax` touches the parabola `x^2= 4by` the locus of P is the straight line

Answer

To solve the problem, we need to find the locus of the point \( P(h, k) \) such that the perpendicular drawn from \( P \) on its polar with respect to the parabola \( y^2 = 4ax \) touches the parabola \( x^2 = 4by \).

### Step 1: Identify the polar of point \( P(h, k) \) with respect to the parabola \( y^2 = 4ax \).

The equation of the polar of a point \( P(h, k) \) with respect to the parabola \( y^2 = 4ax \) is given by:
\[ ky = 2a(x + h) \]
Rearranging gives:
\[ ky - 2ax - 2ah = 0 \]

### Step 2: Find the equation of the perpendicular from point \( P(h, k) \).

The slope of the line \( ky = 2ax + 2ah \) is \( \frac{2a}{k} \). The slope of the perpendicular line will be the negative reciprocal:
\[ \text{slope of perpendicular} = -\frac{k}{2a} \]
Using point-slope form, the equation of the perpendicular line through \( P(h, k) \) is:
\[ y - k = -\frac{k}{2a}(x - h) \]
Rearranging gives:
\[ y = -\frac{k}{2a}x + \left(k + \frac{kh}{2a}\right) \]

### Step 3: Find the condition for this line to touch the parabola \( x^2 = 4by \).

The equation of the parabola \( x^2 = 4by \) can be rearranged to:
\[ 4by - x^2 = 0 \]
For the line to touch the parabola, the discriminant of the resulting quadratic equation must be zero. Substitute \( y \) from the perpendicular line into the parabola's equation:
\[ x^2 = 4b\left(-\frac{k}{2a}x + \left(k + \frac{kh}{2a}\right)\right) \]
This simplifies to:
\[ x^2 + 2b\frac{k}{2a}x - 4b\left(k + \frac{kh}{2a}\right) = 0 \]
The discriminant \( D \) of this quadratic equation must be zero for tangency:
\[ D = \left(2b\frac{k}{2a}\right)^2 - 4\left(-4b\left(k + \frac{kh}{2a}\right)\right) = 0 \]
Simplifying this gives:
\[ \frac{b^2k^2}{4a^2} + 16b\left(k + \frac{kh}{2a}\right) = 0 \]

### Step 4: Solve for the locus of point \( P(h, k) \).

Rearranging the equation obtained from the discriminant condition leads to:
\[ b^2k^2 + 64ab\left(k + \frac{kh}{2a}\right) = 0 \]
Factoring out \( k \):
\[ k\left(b^2k + 64ab + 32bh\right) = 0 \]
This gives two cases: \( k = 0 \) or \( b^2k + 64ab + 32bh = 0 \).

The first case \( k = 0 \) implies the locus is the x-axis. The second case can be rearranged to express \( h \) in terms of \( k \):
\[ h = -\frac{b^2k + 64ab}{32b} \]
This indicates a linear relationship between \( h \) and \( k \), confirming that the locus of point \( P \) is a straight line.

### Final Result:
The locus of point \( P(h, k) \) is a straight line.

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