Let a line L pass through the point of intersection of the lines $bx+10y-8=0$ and $2x-3y=0$, $\mathrm{b} \in \mathrm{R}-\left\{\frac{4}{3}\right\}$. If the line L also passes through the point $(1,1)$ and touches the circle $17\left(x^{2}+y^{2}\right)=16$, then the eccentricity of the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{b^{2}}=1$ is:

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Ellipse

Let a line L pass through the point of intersection of the lines

b x + 10 y - 8 = 0

and

2 x - 3 y = 0

b \in R - {\frac{4}{3}}

. If the line 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

is:

Answer: (B)

The line through the intersection is

(b x + 10 y - 8) + λ (2 x - 3 y) = 0

.

Passing through

(1, 1)

(b + 2) + λ (- 1) = 0

, so

λ = b + 2

.

Line equation:

(3 b + 4) x - (3 b - 4) y - 8 = 0

For tangency to circle

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

\frac{8}{( 3 b + 4 ) ^{2} + ( 3 b - 4 ) ^{2}} = \frac{4}{17}

Squaring and simplifying:

18 b^{2} = 36

, so

b^{2} = 2

For the ellipse

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

(since

a^{2} = 5, b^{2} = 2

e = 1 - \frac{2}{5} = \frac{3}{5}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option