Question bank/JEE Main/Mathematics

JEE MainMathematicsModerateConic Sections

Let 0<θ<π20 < \theta < \frac{\pi}{2}. If the eccentricity of the
hyperbolax2cos⁡2θ−y2sin⁡2θ=1 \frac{x^{2}}{\cos^{2}\theta} - \frac{y^{2}}{\sin^{2}\theta} = 1\ is
greater than 2, then the length of its latus rectum lies in the
interval

Show answer and solution
x2cos⁡2θ−y2sin⁡2θ=1\frac{x^{2}}{\cos^{2}\theta} - \frac{y^{2}}{\sin^{2}\theta} = 1
∴ e > 2 (given)
e2>4⇒1+sin⁡2θcos⁡2θ>4e^{2} > 4 \Rightarrow 1 + \frac{\sin^{2}\theta}{\cos^{2}\theta} > 4
⇒1+tan⁡2θ>4⇒tan⁡2θ>3\Rightarrow 1 + \tan^{2}\theta > 4 \Rightarrow \tan^{2}\theta > 3
∵θ∈(0,π2) hence θ∈(π3,π2)\because\theta \in \left( 0,\frac{\pi}{2} \right)\ hence\ \theta \in \left( \frac{\pi}{3},\frac{\pi}{2} \right)
Latus rectum
=2sin⁡2θcosθ=2tanθsinθ= 2\frac{\sin^{2}\theta}{cos\theta} = 2tan\theta sin\theta
∵for θ∈(π3,π2)\because for\ \theta \in \left( \frac{\pi}{3},\frac{\pi}{2} \right),
2tanθsinθ2tan\theta sin\theta is
An increasing function
Hence latus rectum ∈(3, ∞)

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans