Let the common tangents to the curves 4(x2 + y2) = 9 and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then
\(\frac{1}{e^2}\) is equal to
Let the common tangents to the curves 4(x2 + y2) = 9 and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then
\(\frac{1}{e^2}\) is equal to
Correct Answer: 4
Solution and Explanation
The correct answer is 4
Let y = mx + c is the common tangent
So, \(c=\frac{1}{m}=±\frac{3}{2}\sqrt{1+m^2}⇒m^2=\frac{1}{3}\)
So equation of common tangents will be
\(y=±\frac{1}{\sqrt3}x±\sqrt3\)
which intersects at Q(–3, 0)
Major axis and minor axis of ellipse are 12 and 6. So eccentricity
\(e^2=1−\frac{1}{4}=\frac{3}{4}\)
and length of latus rectum \(=\frac{2b^2}{a}=3\)
Therefore ,
\(\frac{l}{e^2}=\frac{3}{\frac{3}{4}}=4\)
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When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
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