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Ellipse

Section: Mathematics

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Question : 54 of 117

Marks: +1, -0

Let a line

L

pass through the point of intersection of the lines

b x+10 y-8=0

2 x-3 y=0, b \in R - \{ \; \frac{4}{3} \}

L

also passes through the point

(1,1)

and touches the circle

17(x^2+y^2)=16

, then the eccentricity of the ellipse

\; \frac{x^2}{5} + \; \frac{y^2}{b^2} = 1

[29-Jul-2022-Shift-1]

$\frac{2}{\sqrt{5}}$
$\sqrt{\; \frac{3}{5}}$
$\frac{1}{\sqrt{5}}$
$\sqrt{\; \frac{2}{5}}$

Solution:

L_1: b x+10 y-8=0, L_2: 2 x-3 y=0

then

L:(b x+10 y-8)+\lambda(2 x-3 y)=0

\because

It passes through

(1,1)

\therefore b+2-\lambda=0 \Rightarrow \lambda=b+2

and touches the circle

x^2+y^2=\; \frac{16}{17}

\left| \; \frac{8^2}{(2\lambda+b)^2+(10-3\lambda)^2} \right| = \; \frac{16}{17}

\Rightarrow 4\lambda^2+b^2+4b\lambda+100+9\lambda^2-60\lambda=68

\Rightarrow 13(b+2)^2+b^2+4b(b+2)-60(b+2)+32=0

\Rightarrow 18b^2=36

\therefore b^2=2

\therefore

Eccentricity of ellipse :

\; \frac{x^2}{5} + \; \frac{y^2}{b^2} = 1

is

\therefore e=\sqrt{1-\; \frac{2}{5}} = \sqrt{\; \frac{3}{5}}

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