Jee Main Math Class 11 Hyperbola JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (d)

Explanation: $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , then $e^2 = \frac{11}{14} l($ I be the length of $\text{LR})$ $\Rightarrow a^2 + b^2 = \frac{11}{7} b^2 a \dots \text{ (i) }$ and $e'^2 = \frac{11}{8} l'$ (I' be the length of LR of conjugate hyperbola) $\Rightarrow a^2 + b^2 = \frac{11}{4} a^2 b$ By (i) and (ii) $7 a = 4 b$ then by (i) $\frac{16}{49} b^2 + b^2 = \frac{11}{7} b^2 \cdot \frac{4b}{7}$ $\Rightarrow 44 b = 65 \text{ and } 77 a = 65$ $\therefore 77 a + 44 b = 130$

Correct Answer: (a)

Explanation: Since, (3,3) lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$ ....(1) Now, normal at (3,3) is $y - 3 = -\frac{a^{2}}{b^{2}}(x - 3)$ , which passes through $(9,0) \Rightarrow b^{2} = 2 a^{2}$ ....(2) So, $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 3$ Also, $a^{2} = \frac{9}{2}$ (from (i) & (ii)) Thus, $(a^{2}, e^{2}) = \left(\frac{9}{2}, 3\right)$

Correct Answer: 182

Explanation: LR subtends $60^{\circ}$ at centre  $\Rightarrow \tan 30^{\circ} = \;\frac{\frac{b^2}{a}}{a e} = \;\frac{b^2}{a^2 e}$ $=\;\frac{1}{\sqrt{3}}$ $\Rightarrow e = \;\frac{\sqrt{3} b^2}{9}$ Also, $e^2 = 1 + \;\frac{b^2}{9}$ $\Rightarrow 1 + \;\frac{b^2}{9} = \;\frac{3 b^4}{81}$ $\Rightarrow b^4 = 3 b^2 + 27$ $\Rightarrow b^4 - 3 b^2 - 27 = 0$ $\Rightarrow b^2 = \;\frac{3}{2}(1+\sqrt{13})$ $\Rightarrow \ell = 3, m = 2, n = 13$ $\Rightarrow \ell^2 + m^2 + n^2 = 182$

Correct Answer: (b)

Explanation: Ellipse $:\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$ $\therefore$ foci $=(\pm 1,0)$ for hyperbola, given $2 a=\sqrt{2} \Rightarrow a=\frac{1}{\sqrt{2}}$ $\therefore$ hyperbola will be $\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{b^{2}}=1$ eccentricity $=\sqrt{1+2 b^{2}}$ $\therefore$ foci $=\left(\pm \sqrt{\frac{1+2 b^{2}}{2}}, 0 \right)$ Ellipse and hyperbola have same foci $\Rightarrow \sqrt{\frac{1+2 b^{2}}{2}}=1$ $\Rightarrow b^{2}=\frac{1}{2}$ $\therefore$ Equation of hyperbola $:\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{\frac{1}{2}}=1$ $\Rightarrow x^{2}-y^{2}=\frac{1}{2}$ Clearly $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}} \right)$ does not lie on it.

Correct Answer: 2

Explanation: $E : \; \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \rightarrow e$ $H : x^2 - y^2 = \frac{1}{2} \Rightarrow e' = \sqrt{2}$ $e = \frac{1}{\sqrt{2}}$ $\because e^2 = \frac{1}{2}$ $1 - \frac{b^2}{a^2} = \frac{1}{2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{2}$ $a^2 = 2 b^2$ $E \& H$ are at right angle they are confocal Focus of Hyperbola $=$ focus of ellipse $\left( \pm \frac{1}{\sqrt{2}} \cdot \sqrt{2}, 0\right) = \left( \pm \frac{a}{\sqrt{2}}, 0\right)$ $a = \sqrt{2}$ $\because a^2 = 2 b^2 \Rightarrow b^2 = 1$ Length of LR $= \frac{2 b^2}{a} = \frac{2(1)}{\sqrt{2}}$ $= \sqrt{2}$ Square of LR $=2$