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  • The coordinates of all points P on 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%5E2%7D%7Ba%5E2%7Da

The coordinates of all points P on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} for which the area of the triangle PON is maximum, where O denotes the origin and N, the foot of the perpendicular from O to the tangent at P.

Option: 1

\mathrm{\left( \pm \frac{b^2}{a^2+b^2}, \pm \frac{a^2}{a^2+b^2}\right)}


Option: 2

\mathrm{\left( \pm \frac{a^2}{\sqrt{a^2+b^2}}, \pm \frac{b^2}{\sqrt{a^2+b^2}}\right)}


Option: 3

\mathrm{\left( \pm \frac{a}{\sqrt{a^2+b^2}}, \pm \frac{b}{\sqrt{a^2+b^2}}\right)}


Option: 4

\mathrm{\left( \pm \frac{a^2}{a^2+b^2}, \pm \frac{b^2}{b^2+a^2}\right)}


Answers (1)

best_answer

\mathrm{\text { Equation of tangent at }(a \cos \theta, b \sin \theta) \text { on the ellipse is }}

\mathrm{\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1}             ....[1]

length of perpendicular,ON , from origin O  to (1) is 

\mathrm{O N=\frac{a b}{\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}}}

Referring figure, NP = OR.
OR is the length of perpendicular drawn from O to normal line to ellipse through P. Equation of normal is

\mathrm{b y-a x \tan \theta+a^2 \sin \theta-b^2 \sin \theta=0}

\mathrm{O R=\frac{\left(a^2-b^2\right) \sin \theta \cos \theta}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}}

Hence, area of ?ONP,

\mathrm{A=\frac{1}{2} O N \cdot N P=\frac{1}{2} \frac{a b\left(a^2-b^2\right) \tan \theta}{\left(b^2+a^2 \tan ^2 \theta\right)}}

\mathrm{\text { For } A \text { to be maximum } \frac{d A}{d \theta}=0 \text {. }}

\mathrm{\begin{aligned} & \text { or } \sec ^2 \theta\left(b^2-a^2 \tan ^2 \theta\right)=0 \\ & \tan \theta= \pm \frac{b}{a} \\ & \cos \theta= \pm \frac{a}{\sqrt{a^2+b^2}} \\ & \sin \theta= \pm \frac{b}{\sqrt{a^2+b^2}} \end{aligned}}

\mathrm{\therefore \quad \text { Required points are }\left( \pm \frac{a^2}{\sqrt{a^2+b^2}}, \pm \frac{b^2}{\sqrt{a^2+b^2}}\right)}

Posted by

shivangi.shekhar

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