Let a line L pass through point of intersection of lines b x+10 y-8=0 and 2 x-3 y=0 , b ∈ R -left{

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10-2. Parabola, Ellipse, Hyperbola

hard

Let a line $L$ pass through the point of intersection of the lines $b x+10 y-8=0$ and $2 x-3 y=0$, $b \in R -\left\{\frac{4}{3}\right\}$. 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.

$\frac{2}{\sqrt{5}}$

$\sqrt{\frac{3}{5}}$

$\frac{1}{\sqrt{5}}$

$\sqrt{\frac{2}{5}}$

(JEE MAIN-2022)

Solution

Line is passing through intersection of $b x+10 y-8=0$ and $2 x-3 y=0$ is $(b x+10 y-8)+\lambda(2 x-3 y)=0$. As line is passing through $(1,1)$ so $\lambda=b+2$

Now line $(3 b+4) x-(3 b-4) y-8=0$ is tangent to circle $17\left(x^{2}+y^{2}\right)=16$

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

$b^{2}=2 \Rightarrow e=\sqrt{\frac{3}{5}}$

Standard 11

Mathematics

The latus rectum of an ellipse is $10$ and the minor axis is equal to the distance between the foci. The equation of the ellipse is

easy

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hard

(JEE MAIN-2013)

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medium

(JEE MAIN-2013)

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normal

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