Test Index

Ellipse

Section: Mathematics

© examsnet.com

Question : 83 of 117

Marks: +1, -0

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse,

\frac{x^{2}}{4}+\frac{y^{2}}{2}=1

from any of its foci ?

[6 Sep 2020 Shift 1]

$(-1,\sqrt{3})$
$(-1,\sqrt{2})$
$(-2,\sqrt{3})$
(1,2)

Solution:

Let foot of perpendicular is (h, k)

\frac{x^{2}}{4}+\frac{y^{2}}{2}=1

(Given)

a=2, b=\sqrt{2}

,

e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}

∴ Focus

(ae,0)=(\sqrt{2},0)

Equation of tangent,

y=mx+\sqrt{a^{2}m^{2}+b^{2}}

y=mx+\sqrt{4m^{2}+2}

Passes through (h, k)

(k-mh)^{2}=4m^{2}+2

....(1)

line perpendicular to tangent will have slope

-\frac{1}{m}

y-0=-\frac{1}{m}(x-\sqrt{2})

my=-x+\sqrt{2}

(h+mk)^{2}=2

.....(2)

Add equation (1) and (2)

k^{2}(1+m^{2})+h^{2}(1+m^{2})=4(1+m^{2})

h^{2}+k^{2}=4

x^{2}+y^{2}=4

(Auxiliary circle)

\therefore(-1,\sqrt{3})

lies on the locus.

© examsnet.com

Go to Question:

Prev Question

Next Question