JEE Main 2022 · 29 Jul · Shift 1 · Mathematics Q8

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MathematicsSingle correct+4 / -1Question 8 of 90

Let a line L pass through the point of intersection of the lines $bx+10y-8=0$ and $2x-3y=0$ , $b\in\mathbf R-\left\{\frac43\right\}$ . 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 $\dfrac{x^2}5+\dfrac{y^2}{b^2}=1$ is :

\dfrac2{\sqrt5}

\sqrt{\dfrac35}

\dfrac1{\sqrt5}

\sqrt{\dfrac25}

Answer and worked solution

Official answer: B

Family. $L:\ bx+10y-8+\lambda(2x-3y)=0$ through (1, 1) gives $\lambda=b+2$ , so $L:\ (3b+4)x+(4-3b)y=8$ .

Tangency. $\dfrac{8}{\sqrt{(3b+4)^2+(4-3b)^2}}=\dfrac4{\sqrt{17}}$ gives $18b^2+32=68$ , $b^2=2$ .

Eccentricity $e^2=1-\frac25=\frac35$ .

(B)

Topics:Family of lines and pair of lines Tangents and normals Ellipse

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JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 8