Let 'd' be the perpendicular distance from the centre of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ to the tangent drawn at at point 'P' on the ellipse. If $F_1$ and $F_2$ are the two foci of the ellipse , then show that $(PF_1-PF_2)^2=4a^2(1-\frac{b^2}{d^2})$.
My approach is as follows:
$P{F_1} = \sqrt {{{\left( {{x_1} - ae} \right)}^2} + {y_1}^2} ;P{F_2} = \sqrt {{{\left( {{x_1} + ae} \right)}^2} + {y_1}^2} $
${\left( {P{F_1} - P{F_2}} \right)^2} = {\left( {{x_1} - ae} \right)^2} + {y_1}^2 + \left( {{{\left( {{x_1} + ae} \right)}^2} + {y_1}^2} \right) - 2\sqrt {{{\left( {{x_1} - ae} \right)}^2} + {y_1}^2} \times \sqrt {{{\left( {{x_1} + ae} \right)}^2} + {y_1}^2} $
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2}{e^2}} \right) + 2{y_1}^2 - 2\sqrt {{x_1}^2 + {a^2}{e^2} + {y_1}^2 - 2ae{x_1}} \times \sqrt {{x_1}^2 + {a^2}{e^2} + {y_1}^2 + 2ae{x_1}} $
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2}{e^2}} \right) + 2{y_1}^2 - 2\sqrt {{{\left( {{x_1}^2 + {a^2}{e^2} + {y_1}^2} \right)}^2} - 4{a^2}{e^2}{x_1}^2} $
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2}{e^2}} \right) + 2{y_1}^2 - 2\sqrt {\left( {{x_1}^4 + {a^4}{e^4} + {y_1}^4 + 2{a^2}{e^2}{x_1}^2 + 2{a^2}{e^2}{y_1}^2 + 2{x_1}^2{y_1}^2} \right) - 4{a^2}{e^2}{x_1}^2} $
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2}{e^2}} \right) + 2{y_1}^2 - 2\sqrt {\left( {{x_1}^4 + {a^4}{e^4} + {y_1}^4 - 2{a^2}{e^2}{x_1}^2 + 2{a^2}{e^2}{y_1}^2 + 2{x_1}^2{y_1}^2} \right)} $
${a^2}{e^2} = {a^2} - {b^2}$
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2} - {b^2}} \right) + 2{y_1}^2 - 2\sqrt {\left( {{x_1}^4 + {{\left( {{a^2} - {b^2}} \right)}^2} + {y_1}^4 - 2\left( {{a^2} - {b^2}} \right){x_1}^2 + 2\left( {{a^2} - {b^2}} \right){y_1}^2 + 2{x_1}^2{y_1}^2} \right)} $
${\left( {P{F_1} - P{F_2}} \right)^2} = 2\left( {{x_1}^2 + {a^2} - {b^2}} \right) + 2{y_1}^2 - 2\sqrt {\left( {{x_1}^4 + {a^4} + {b^4} - 2{a^2}{b^2} + {y_1}^4 - 2\left( {{a^2} - {b^2}} \right){x_1}^2 + 2\left( {{a^2} - {b^2}} \right){y_1}^2 + 2{x_1}^2{y_1}^2} \right)}$
$T = \sqrt {\left( {{x_1}^4 + {a^4} + {b^4} - 2{a^2}{b^2} + {y_1}^4 - 2\left( {{a^2} - {b^2}} \right){x_1}^2 + 2\left( {{a^2} - {b^2}} \right){y_1}^2 + 2{x_1}^2{y_1}^2} \right)} $
$T = \sqrt {\left( {{x_1}^4 + {a^4} + {b^4} + {y_1}^4 - 2{a^2}{b^2} - 2{a^2}{x_1}^2 + 2{b^2}{x_1}^2 + 2{a^2}{y_1}^2 - 2{b^2}{y_1}^2 + 2{x_1}^2{y_1}^2} \right)} $
I need to make the values a perfect square but not able to do so.
Tip : Use polar coordinates for length-related properties of conics.
As shown in comment below the question, $$(PF_1-PF_2)^2=4a^2-4PF_1\cdot PF_2$$
So we just need to evaluate $PF_1\cdot PF_2$
Let $P=(a\cos \theta, b\sin \theta)$. Equation of tangent would be $$bx\cos \theta+ay\sin \theta-ab=0$$ $$\Rightarrow d^2 = \frac{a^2b^2}{b^2\cos^2 \theta + a^2\sin^2\theta} \tag{$\star$}$$
Using $\text{focal distance} =e \times \text{directrix distance}$, $$PF_1=e\left(\frac{a}{e}+a\cos \theta\right) \quad , \quad PF_2=e\left(\frac{a}{e}-a\cos \theta\right)$$ $$PF_1=a+ae\cos \theta \quad , \quad PF_2=a-ae\cos \theta$$ $$PF_1 \cdot PF_2 = a^2 - a^2e^2\cos^2 \theta$$ $$=a^2-(a^2-b^2)\cos^2 \theta=a^2\sin^2\theta + b^2\cos^2\theta$$
From $(\star)$ we get that $$PF_1 \cdot PF_2 = \frac{a^2b^2}{d^2}$$
which is as required.