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The locus of the foot of the perpendicular from a focus to a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is the circle $x^2 + y^2 = a^2$

I am able to solve it using tangent as $y=mx±\sqrt{a^2m^2+b^2}$ but my student wants me to solve it using $x=a\cos \theta$ and $y=b\sin \theta$. I am not able to solve it.

Ng Chung Tak
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Samar Imam Zaidi
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  • The form of equation of a line that you’ve used successfully doesn’t include vertical tangents, but no doubt you’ve dealt with that in your proof. – amd Jan 10 '18 at 20:45

1 Answers1

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Using this, the equation at $P(a\cos t,b\sin t)$

is $$xb\cos t+y a\sin t-ab=0$$

Method$\#1:$

Now if $R(h,k)$ is the foot of the perpendicular from focus $A(ae,0)$

$$hb\cos t+k a\sin t-ab=0\ \ \ \ (1)$$

and as $PR$ will be perpendicular to the tangent, form the second equation $$\dfrac{0-k}{ae-h}\cdot-\dfrac{b\cos t}{a\sin t}=-1\ \ \ \ (2)$$

Solve for $h,k$ and eliminate $t$

Method$\#2:$

As the gradient of the tangent is $$-\dfrac{b\cos t}{a\sin t},$$ the equation of the perpendicular from focus $A(ae,0)$ will be

$$\dfrac{a\sin t}{b\cos t}=\dfrac{y-0}{x-ae}$$

Find the intersection of the perpendicular with the tangent and eliminate $t$

lab bhattacharjee
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