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For a standard ellipse, a chord subtends an angle of $90^{\circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:

$$\frac xa\space\cos{A+B\over 2} + \frac yb\space\sin{A+B\over 2} = \cos{A-B\over 2}$$ Since an angle of $90^{\circ}$ is subtended at the centre, I get the condition that $\tan A\tan B = -\dfrac {a^2}{ b^2}$

By imposing the condition that FoP is perpendicular to the chord, I get $\dfrac {ax}b\tan{A+B\over 2} = y$.

Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.

najayaz
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Sat D
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2 Answers2

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Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.

Sat D
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Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at $$ A=\left({ab\over\sqrt{m^2a^2+b^2}},{mab\over\sqrt{m^2a^2+b^2}}\right). $$ A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at $$ B=\left({mab\over\sqrt{a^2+m^2b^2}},{-ab\over\sqrt{a^2+m^2b^2}}\right). $$ You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.

Intelligenti pauca
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