Answer to Question #104114 in Analytic Geometry for Deepak Rana

Question #104114
The normals at any point P of the ellipsoid x
2
9 +
y
2
4 +z
2 = 1 meet the coordinate
planes in Q1,Q2,Q3, respectively. Show that PQ1 : PQ2 : PQ3 :: 9 : 4 : 1.
1
Expert's answer
2020-03-03T10:58:24-0500

As per the given condition in the question,

The equation of the ellipse,

x29+y24+z2=1\dfrac{x^2}{9}+\dfrac{y^2}{4}+z^2=1

now compare the above the general equation of the ellipse,

x2a2+y2b2+z2c2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1

Let the parametric point P(asinθcosϕ,bcosθsinϕ,ccosθ)P(a\sin\theta\cos\phi, b\cos\theta\sin\phi, c\cos\theta)

a=3, b=2, c=1

now putting (asinθcosϕ,0,0)(a\sin\theta\cos\phi, 0, 0),

We know from the general equation,

we know that from the general equation,

x=aa2b2c2=94x=-\dfrac{a}{a^2-b^2-c^2}=\dfrac{9}{-4}

similarly y=ba2b2c2=44y=-\dfrac{b}{a^2-b^2-c^2}=\dfrac{4}{-4}

similarly z=ca2b2c2=14z=\dfrac{c}{a^2-b^2-c^2}=\dfrac{-1}{4}

So, distance PQ1=(x0)2+(00)2+(00)2=94PQ_1=\sqrt{(x-0)^2+(0-0)^2+(0-0)^2}=\dfrac{9}{4}

Similarly PQ2=44PQ_2=\dfrac{4}{4}

PQ3=14PQ_3=\dfrac{1}{4}

Hence the required ratio is 9:4:1



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Comments

Assignment Expert
04.03.20, 15:35

Dear Deepak, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Deepak
04.03.20, 12:33

Thanks for your help

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