JEE MAIN Hyperbola Previous Year Questions (PYQs) – Page 1 of 3
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Consider a hyperbola $\text{H}$ having centre at the origin and foci on the x-axis.
Let $C_1$ be the circle touching the hyperbola $\text{H}$ and having the centre at the origin.
Let $C_2$ be the circle touching the hyperbola $\text{H}$ at its vertex and having the centre at one of its foci.
If areas (in sq units) of $C_1$ and $C_2$ are $36\pi$ and $4\pi$, respectively,
then the length (in units) of latus rectum of $\text{H}$ is:
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Solution
Let the matrix $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$ satisfy $A^n=A^{n-2}+A^2-I$ for $n \geqslant 3$. Then the sum of all the elements of $\mathrm{A}^{50}$ is :
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Solution
Given $A^n=A^{n-2}+A^2-I$ for $n\ge 3$
Let $S_n$ = sum of all elements of $A^n$
Taking sum on both sides:
$S_n=S_{n-2}+S_2-3$ (since sum of $I=3$)
Compute base values:
$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$
⇒ $S_1=3$
$A^2=\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$
⇒ $S_2=5$
So recurrence:
$S_n=S_{n-2}+2$
Even terms:
$S_2=5,\ S_4=7,\ S_6=9,\dots$
⇒ $S_{2k}=5+(k-1)\cdot2=2k+3$
For $n=50$, $k=25$:
$S_{50}=2(25)+3=53$
Let $S_n$ = sum of all elements of $A^n$
Taking sum on both sides:
$S_n=S_{n-2}+S_2-3$ (since sum of $I=3$)
Compute base values:
$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$
⇒ $S_1=3$
$A^2=\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$
⇒ $S_2=5$
So recurrence:
$S_n=S_{n-2}+2$
Even terms:
$S_2=5,\ S_4=7,\ S_6=9,\dots$
⇒ $S_{2k}=5+(k-1)\cdot2=2k+3$
For $n=50$, $k=25$:
$S_{50}=2(25)+3=53$
Let $S = \{(x, y) \in \mathbb{R}^2 : \dfrac{y^2}{1 + r} - \dfrac{x^2}{1 - r} = 1 ; \, r \neq \pm 1 \}$.
Then $S$ represents :
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Solution
If the line $x-1=0$ is a directrix of the hyperbola $kx^{2}-y^{2}=6$, then the hyperbola passes through the point:
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Solution
If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
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Solution
$y = \frac{1 - \alpha x}{2}$
Put this in equation of hyperbola
$x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9$
$\Rightarrow (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$
Line does not intersect hyperbola
$\Rightarrow D < 0$
$\Rightarrow \alpha^2 < \frac{5}{9}$
$\Rightarrow \alpha \in \left(-\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3}\right)$
$\Rightarrow \frac{\sqrt{5}}{3} \approx 0.74$
$\Rightarrow \alpha = 0.8$ (possible value)
If a hyperbola has length of its conjugate axis equal to $5$ and the distance between its foci is $13$, then the eccentricity of the hyperbola is :
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Solution
Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to :
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Solution
For hyperbola, latus rectum length $=\dfrac{2b^2}{a}=8\Rightarrow b^2=4a$. Since $P(10,2\sqrt{15})$ lies on it, $\dfrac{100}{a^2}-\dfrac{60}{b^2}=1$. Using $b^2=4a$, we get $a=5$, so $b^2=20$ and $c^2=a^2+b^2=45$. Hence $SS'=2c=6\sqrt5$. Area of $\Delta PSS' = \dfrac12\cdot 6\sqrt5\cdot 2\sqrt{15}=30\sqrt3$. Therefore square of area $=(30\sqrt3)^2=2700$.
If the vertices of a hyperbola are at $(-2,0)$ and $(2,0)$ and one of its foci is at $(-3,0)$, then which one of the following points does not lie on this hyperbola?
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Solution
Let $PQ$ be a chord of the hyperbola
$\frac{x^2}{4} - \frac{y^2}{b^2} = 1$,
perpendicular to the x-axis such that triangle $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of triangle $OPQ$ is:
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Solution
$e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$
$\Rightarrow b = 8$
Hyperbola
$\frac{x^2}{4} - \frac{y^2}{64} = 1$
$PM/OM = \tan30^\circ$
$\Rightarrow \frac{2\sqrt{2}\tan\theta}{2\sec\theta} = \frac{1}{\sqrt{3}}$
$\Rightarrow \sin\theta = \frac{1}{\sqrt{6}}$
Area
$= 2 \times \frac{1}{2} \times OM \times MP$
$= 2\sec\theta \cdot 2\sqrt{2}\tan\theta$
$= 4\sqrt{2} \cdot \frac{\sin\theta}{\cos^2\theta}$
$= 4\sqrt{2} \cdot \frac{1}{\sqrt{6}\left(1 - \frac{1}{6}\right)}$
$= \frac{8\sqrt{3}}{5}$
If the function
$f(x)=\dfrac{e^x\left(e^{\tan x}-1\right)+\ln(\sec x+\tan x)-x}{\tan x-x}$
is continuous at $x=0$, then the value of $f(0)$ is equal to:
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Solution
$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}-e^x+\ln(\sec x+\tan x)-x}{\tan x-x}$
Applying L’Hospital rule
$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}\sec^2x-e^x+\sec x-1}{\sec^2x-1}$
$=\lim_{x\to0}\dfrac{e^{\tan x}(\sec^2x-1)+(e^{\tan x}-e^x)+\sec x-1}{\tan^2x}$
$=\lim_{x\to0}\left(\dfrac{e^{\tan x}(e^{\tan x-x}-1)}{\tan^2x}+\dfrac{1}{\sec x+1}\right)$
$\Rightarrow f(0)=1+0+\frac{1}{2}=\frac{3}{2}$
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