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Through point $P(2,2)$ is drawn a chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16} = 1$ such that it intersect

the ellipse at $A$ and $B$. then maximum value of $PA\cdot PB$ is

Attempt: point $P(2,2)$ lie inside the ellipse $\frac{x^2}{25}+\frac{y^2}{16} = 1$

let line through $P(2,2)$ makes an angle of $\theta $ with x-axis, then $\frac{x-2}{\cos \theta} = \frac{y-2}{\sin \theta} = r$

so $x=r\cos \theta +2$ and $y=r\sin \theta+2$

so $\frac{(r\cos \theta+2)^2}{25}+\frac{(r\sin \theta+2)^2}{16} = 1\Rightarrow 16(r\cos \theta+2)^2+25(r\sin \theta+2)^2 = 25\cdot 16$

so $r^2(16+9\sin^2 \theta)+r(64\cos \theta+100\sin \theta)+164-400=0$

so $ r_{1}\cdot r_{2} = \frac{236}{16+9\sin^2 \theta}$, where $PA=r_{1}$ and $PB=r_{2}$

so $\max(r_{1}r_{2}) = \frac{236}{9}$ when $\sin^2 \theta = 0$

is my try is right, if any error then could some help me to solve it , thanks

Nij
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DXT
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1 Answers1

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Your approach would have been correct if the point lied outside the ellipse but since it lies inside the ellipse your method only calculates the square of one part of the chord on which the point lies because since an ellipse is symmetrical you can draw two equal chords through a point.To solve this correctly remember that the diameter of a ellipse through a point is largest among all chords through that point.You must be knowing equation of diameter of an ellipse.

catechol
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