If the product of perpendiculars from the foci upon the polar of $P$ be constant and equal to $c^{2}$ , prove that the locus of $P$ is the ellipse $b^{2} x^{2} (c^{2} + a^{2} e^{2}) + c^{2} a^{4} y^{2} = a^{4} b^{4}$ .

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Solution

Step 1: Determine the polar of $P$
It is given that the product of perpendiculars from the foci upon the polar of $P$ is $c^{2}$ .
We know that the equation of an ellipse whose center is at origin is,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
And, the coordinates of foci are $(a e, 0)$ and $(- a e, 0)$ .
Let, the coordinates of $P$ be $(h, k)$ .
Polar of $P$ is,
$\begin{array}{rcl} \frac{x h}{a^{2}} + \frac{y k}{b^{2}} & = & 1 \\ \Rightarrow & b^{2} h x + a^{2} k y - a^{2} b^{2} = 0 . . (i) \end{array}$
Step 2: Calculate the product of perpendiculars from the foci
Let, $P_{1}$ and $P_{2}$ be the lengths of the perpendicular line represented by equation $(i)$ from $(a e, 0)$ and $(- a e, 0)$ on $(i)$ .
Then, $P_{1} = \frac{(b^{2} h a e - a^{2} b^{2})}{\sqrt{(b^{4} h^{2} + a^{4} k^{2})}}$
And, $P_{2} = \frac{(- a^{2} b^{2} - b^{2} h a e)}{\sqrt{(b^{4} h^{4} + a^{4} k^{2})}}$
Now, the product of $P_{1}$ and $P_{2}$ can be calculated as,
$P_{1} \times P_{2} = \frac{(b^{2} h a e - a^{2} b^{2}) (- a^{2} b^{2} - b^{2} h a e)}{(\sqrt{b^{4} h^{2} + a^{4} k^{2}}) (\sqrt{b^{4} h^{2} + a^{4} k^{2}})}$
$\Rightarrow$ $P_{1} \times P_{2} = \frac{- (b^{2} h a e - a^{2} b^{2}) (b^{2} h a e + a^{2} b^{2})}{(b^{4} h^{2} + a^{4} k^{2})}$
$\Rightarrow$ $P_{1} \times P_{2} = \frac{- [{(b^{2} h a e)}^{2} - {(a^{2} b^{2})}^{4}]}{(b^{4} h^{2} + a^{4} k^{2})}$ $[∵ (a + b) (a - b) = a^{2} - b^{2}]$
$\Rightarrow$ $P_{1} \times P_{2} = \frac{- b^{4} h^{2} a^{2} e^{2} + a^{4} b^{4}}{b^{4} h^{2} + a^{4} k^{2}} . . . (ii)$
Step 3: Deduce the given equation
Now, according to the question, $P_{1} \times P_{2} = c^{2}$ . So, the equation $(ii)$ becomes,
$c^{2} = \frac{- b^{4} h^{2} a^{2} e^{2} + a^{4} b^{4}}{b^{4} h^{2} + a^{4} k^{2}}$
$\Rightarrow$ $b^{4} c^{2} h^{2} + a^{4} c^{2} k^{2} = - b^{4} h^{2} a^{2} e^{2} + a^{4} b^{4}$
$\Rightarrow$ $b^{4} c^{2} h^{2} + b^{4} h^{2} a^{2} e^{2} + a^{4} c^{2} k^{2} = a^{4} b^{4}$
$\Rightarrow$ $b^{4} h^{2} (c^{2} + a^{2} e^{2}) + a^{4} c^{2} k^{2} = a^{4} b^{4}$
Now, on replacing $(h, k)$ by $(x, y)$ , we get,
$b^{4} x^{2} (c^{2} + a^{2} e^{2}) + a^{4} c^{2} y^{2} = a^{4} b^{4}$
Which is the required equation.
Hence, it is proved that the locus of $P$ is the given ellipse is $b^{4} x^{2} (c^{2} + a^{2} e^{2}) + a^{4} c^{2} y^{2} = a^{4} b^{4}$ .

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If the product of perpendiculars from the foci upon the polar of $P$ be constant and equal to $c^{2}$ , prove that the locus of $P$ is the ellipse $b^{2} x^{2} (c^{2} + a^{2} e^{2}) + c^{2} a^{4} y^{2} = a^{4} b^{4}$ .