JEE Advance - Mathematics (1995 - No. 7)
Let '$$d$$' be the perpendicular distance from the centre of the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ to the tangent drawn at a point $$P$$ on the ellipse. If $${F_1}$$ and $${F_2}$$ are the two foci of the ellipse, then show that $${\left( {P{F_1} - P{F_2}} \right)^2} = 4{a^2}\left( {1 - {{{b^2}} \over {{d^2}}}} \right)$$.
This question requires knowledge of the properties of an ellipse, including its foci, tangents, and the relationship between the distance from the center to a tangent and the semi-major and semi-minor axes.
The solution involves using the equation of the tangent at a point on the ellipse, finding the distance from the center to the tangent, and then manipulating the expressions using the properties of the foci and the definition of the ellipse.
The problem combines concepts from coordinate geometry and conic sections, demanding a strong understanding of the relationships between different geometric parameters.
The given equation represents a key relationship that needs to be derived, suggesting the need for careful algebraic manipulation and application of relevant formulas.
All of the above.
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