Let a line L pass through the point intersection of the lines \(bx + 10y – 8 = 0\) and \(2x - 3y = 0, \quad b \in \mathbb{R} - \left\{\frac{4}{3}\right\}\). 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 \(\frac{x^2}{5} + \frac{y^2}{5} = 1\) is
\(\frac{2}{\sqrt{5}}\)
\(\frac{\sqrt{3}}{5}\)
\(\frac{1}{\sqrt{5}}\)
\(\frac{\sqrt{2}}{5}\)
The Correct Option is B
Solution and Explanation
\(L_1 :bx + 10y – 8 = 0, L_2 : 2x – 3y = 0\)
then \(L : (bx + 10y – 8) + λ(2x – 3y) = 0\)
Since, It passes through \((1, 1)\)
so, \(b + 2 – λ = 0 ⇒ λ = b + 2\)
and touches the circle \(x^2 + y^2 = \frac{16}{17}\)
\(\left|\frac{82}{(2\lambda + b)^2} + (10 - 3\lambda)^2\right| = \frac{16}{17}\)
\(⇒\) \(4\lambda^2 + b^2 + 4b\lambda + 100 + 9\lambda^2 - 60\lambda = 68\)
\(⇒\) \(13(b+2)^2 + b^2 + 4b(b+2) - 60(b+2) + 32 = 0\)
\(⇒\)\(18b^2=36 ∴b^2=2\)
∴ Eccentricity of ellipse \(\frac{x^2}{5} + \frac{y^2}{b^2} = 1\) is
\(e = \sqrt{1 - \frac{2}{5}}\)
\(e = \frac{\sqrt{3}}{5}\)
So, the correct option is (B): \( \frac{\sqrt{3}}{5}\)
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- Solutions
Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}