JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The locus of the point of intersection of the straight lines, \(tx -2y-3t=0\) \(x - 2ty+ 3 = 0\) \(\left( {t \in R} \right)\), is
- A an ellipse with eccentricity \(\frac{2}{{\sqrt 5 }}\)
- B an ellipse with the length of major axis \(6\)
- C a hyperbola with eccentricity \(\sqrt 5 \)
- D a hyperbola with the length of conjugate axis \(3\)
Answer & Solution
Correct Answer
(D) a hyperbola with the length of conjugate axis \(3\)
Step-by-step Solution
Detailed explanation
Here, \(tx - 2y - 3t = 0\,\ and \,x - 2ty + 3 = 0\) On solving we get; \(y = \frac{{3t}}{{{t^2} - 1}}\,\,\ and \,\,x = \frac{{3{t^2} + 3}}{{{t^2} - 1}}\) Put \(t = \tan \,\theta \) \(\therefore x = - 3\,\sec 2\theta \,\ and \,\,2y = 3\left( { - tan2\theta } \right)\)…
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