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Ellipse

Section: Mathematics

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Question : 8 of 16

Marks: +1, -0

If tangents are drawn to the ellipse

x^2+2 y^2=2

, then the locus of the midpoints of the intercepts made by the tangents between the coordinate axes is

[AP EAPCET 21 May 2025 S1]

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

Solution:

Given, ellipse is

x^2+2 y^2=2

or

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

Now, equation of tangent in parametric form is

\;\frac{x}{\sqrt{2}} \cos\theta + y \sin\;\theta = 1

Now, coordinate of

X

-intercept

=(\sqrt{2} \sec\theta, 0) A

Coordinate of

Y

-intercept

=(0, \csc\theta) B

Let

P(h, k)

be the points which is the mid-points of

A

and

B

.

h\;=\;\frac{\sqrt{2} \sec\theta}{2} \;\text{ and }\; k=\;\frac{\csc\theta}{2}

\sec\theta\;=\sqrt{2} h \;\text{ and }\; \csc\theta = 2 k

\cos\theta\;=\;\frac{1}{\sqrt{2 h}} \;\text{ and }\; \sin\;\theta=\;\frac{1}{2 k}

Using

\sin^2\theta + \cos^2\theta = 1

\;\frac{1}{2 h^2}+\;\frac{1}{4 k^2}=1

Taking locus of

P(h, k)

, we get

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

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