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  • Let d be the perpendicular distance from the centre of the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx

Let d be the perpendicular distance from the centre of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} to the tangent drawn at a point P on the ellipse. If \mathrm{F_1 \ and \ F_2} are the two foci of the ellipse then \mathrm{ \left(P F_1-P F_2\right)^2} is equal to

Option: 1

\mathrm{4 a^2\left(1-\frac{b^2}{d^2}\right)}


Option: 2

\mathrm{4 b^2\left(1-\frac{a^2}{d^2}\right)}


Option: 3

\mathrm{4 a\left(1-\frac{b^2}{d^2}\right)}


Option: 4

\mathrm{2 a^2\left(1-\frac{b^2}{d^2}\right)}


Answers (1)

best_answer

Let a > b and e be the eccentricity of the ellipse.

\mathrm{\text { Then } F_1=(-a e, 0) \text { and } F_2=(a e, 0), P=(a \cos \theta, b \sin \theta)}

\mathrm{\text { Equation of the tangent at } P \text { is } \frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1}

Perpendicular distance from the centre (0, 0) to the

\mathrm{\text { tangent at } P \text { is } d^2=\frac{1}{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}

\mathrm{\text { Now } \begin{aligned} 4 a^2\left(1-\frac{b^2}{d^2}\right)=4 a^2\left[1-b^2\left(\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}\right)\right] \\ =4 a^2 e^2 \cos ^2 \theta \end{aligned}}      ....[1]

\mathrm{\text { But } P F_1=a+a e \cos \theta, P F_2=a-a e \cos \theta}

\mathrm{\therefore \quad\left(P F_1-P F_2\right)^2=(2 a e \cos \theta)^2=4 a^2 e^2 \cos ^2 \theta}                ......[2]

\mathrm{\text { From (1) \& (2), }\left(P F_1-P F_2\right)^2=4 a^2\left(1-\frac{b^2}{d^2}\right)}

Posted by

Ritika Kankaria

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