Test Index

Ellipse

Section: Mathematics

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Question : 53 of 117

Marks: +1, -0

Let

\mathrm{C}

be the largest circle centred at

(2,0)

and inscribed in the ellipse

=\frac{x^2}{36}+\frac{y^2}{16}=1

(1, \alpha)

\mathrm{C}

10 \alpha^2

is equal to____

[24-Jan-2023 Shift 1]

Your Answer:

Solution:

Equation of normal of ellipse

\frac{x^2}{36}+\frac{y^2}{16}=1

at any point

\mathrm{P}(6 \cos \theta, 4 \sin \theta)

is

3 \sec \theta x-2 \csc \theta y=10

this normal is also the normal of the circle passing through the point

(2,0)

So,

6 \sec \theta=10

or

\sin \theta=0

(Not possible)

\cos \theta=\frac{3}{5}

and

\sin \theta=\frac{4}{5}

so point

\mathrm{P}= \left( \frac{18}{5}, \frac{16}{5} \right)

So the largest radius of circle

\sqrt{320}

So the equation of circle

(x-2)^2+y^2=\frac{64}{5}

Passing it through

(1, \alpha)

Then

\alpha^2=\frac{59}{5}

10 \alpha^2=118

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