hyperbola ${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$ = 1 is greater
than 2, then the length of its latus rectum lies in the interval :
If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $\Delta $PTQ is :
tx $-$ 2y $-$ 3t = 0
x $-$ 2ty + 3 = 0 (t $ \in $ R), is :
For some $\theta \in\left(0, \frac{\pi}{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2-y^2 \sec ^2 \theta=8$ be $e_1$ and $l_1$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^2 \sec ^2 \theta+y^2=6$ be $e_2$ and $l_2$, respectively. If $e_1^2=e_2^2\left(\sec ^2 \theta+1\right)$, then $\left(\frac{l_1 l_2}{e_1 e_2}\right) \tan ^2 \theta$ is equal to
Explanation:
Given hyperbola
$ x^2-y^2 \sec ^2 \theta=8 $
Convert it into standard form.
$ \frac{x^2}{8}-\frac{y^2}{8 \cos ^2 \theta}=1 $
Eccentricity: $e_1^2=1+\frac{8 \cos ^2 \theta}{8}=1+\cos ^2 \theta$ latus rectum
$ l_1=\frac{2 \times 8 \cos ^2 \theta}{\sqrt{8}}=4 \sqrt{2} \cos ^2 \theta $
Given ellipse
$ x^2 \sec ^2 \theta+y^2=6 $
Convert into standard form.
$\frac{x^2}{6 \cos ^2 \theta}+\frac{y^2}{6}=1$
Since $\theta \in(0, \pi / 2) \Rightarrow \cos ^2 \theta<1$ so major axis is y -axis. eccentricity:
$ e_2^2=1-\frac{6 \cos ^2 \theta}{6}=\sin ^2 \theta $
Latus rectum
$ l_2=\frac{2 \times 6 \cos ^2 \theta}{\sqrt{6}}=2 \sqrt{6} \cos ^2 \theta $
It is given that
$ e_1^2=e_2^2\left(\sec ^2 \theta+1\right) $
Substituting values of $e_1^2 $ and $ e_2^2$.
$1+\cos ^2 \theta=\sin ^2 \theta\left(\frac{1}{\cos ^2 \theta}+1\right)$
$\Rightarrow $ $1+\cos ^2 \theta=\sin ^2 \theta\left(\frac{1+\cos ^2 \theta}{\cos ^2 \theta}\right)$
$\Rightarrow $ $\tan ^2 \theta=1 \Rightarrow \theta=\frac{\pi}{4}$ as $\theta \in(0, \pi / 2)$
$\Rightarrow $ $\frac{l_1 \cdot l_2}{e_1 \cdot e_2}=\frac{\left(4 \sqrt{2} \cos ^2 \frac{\pi}{4}\right)\left(2 \sqrt{6} \cos ^2 \frac{\pi}{4}\right)}{\sqrt{\left(1+\cos ^2 \frac{\pi}{4}\right)}\left(\sin \frac{\pi}{4}\right)}$
$\therefore $ $\frac{l_1 \cdot l_2}{e_1 \cdot e_2}=\frac{4 \sqrt{2} \times \frac{1}{2} \times 2 \sqrt{6} \times \frac{1}{2}}{\sqrt{\left(1+\frac{1}{2}\right)}\left(\frac{1}{\sqrt{2}}\right)}=\frac{4 \sqrt{3}}{\frac{\sqrt{3}}{2}}=8$
Explanation:
Given:
Transverse axis length: $2a$
Conjugate axis length: $2b$
One focus at $(-5, 0)$
Directrix given by $5x + 9 = 0$
The equations used are as follows:
Relationship between focus and directrix:
The focal length $ae = 5$
The directrix gives $\frac{a}{e} = \frac{9}{5}$
Solving these equations, we get:
$ ae = 5, \quad \frac{a}{e} = \frac{9}{5} $
From $ae = 5$, we have $a = \frac{5}{e}$.
Substituting $a = 3$ and $e = \frac{5}{3}$.
Finding $ b $:
Use the relationship $b^2 = a^2(e^2 - 1)$:
$ \text{Given } a = 3 \quad \text{and } e = \frac{5}{3}, \quad b = 4 $
Equation of the hyperbola:
The standard equation, after substituting values of $a$ and $b$, is:
$ \frac{x^2}{9} - \frac{y^2}{16} = 1 $
Focal distances product calculation:
For any point $(\alpha, 2\sqrt{5})$ that lies on the hyperbola:
$ \frac{\alpha^2}{9} - \frac{(2\sqrt{5})^2}{16} = 1 $
Solving this gives:
$ \alpha^2 = 9 \times \frac{36}{16} $
Product of focal distances $ \mathrm{PF}_1 \cdot \mathrm{PF}_2 $:
$ \mathrm{PF}_1 \cdot \mathrm{PF}_2 = (e\alpha - a)(e\alpha + a) = e^2\alpha^2 - a^2 $
Substituting the values:
$ P = e^2\alpha^2 - a^2 = \frac{25}{9} \cdot 9 \cdot \frac{9}{4} - 9 = \frac{189}{4} $
Finally, to find $4p$:
$ 4p = 4 \times \frac{189}{4} = 189 $
Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its focus at $\mathrm{P}(-3,0)$. If the latus ractum through its other focus subtends a right angle at P and $a^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is _________ .
Explanation:
$\begin{aligned} & a e=3, \tan 45^{\circ}=a e=3 \frac{\frac{b^2}{a}}{6} \Rightarrow \frac{b^2}{a}=6 \ldots \text { (i) } \quad\text{..... (i)}\\ & a \sqrt{\left(1+\frac{b^2}{a^2}\right)}=3 \\ & a^2+b^2=9 \quad\text{..... (ii)}\\ & \Rightarrow a^2-6 a+9=0 \Rightarrow a=3(\sqrt{2}-1) \\ & \Rightarrow a^2 b^2=9(3-2 \sqrt{2}) \cdot 6 \cdot 3(\sqrt{2}-1) \\ & \quad=162(5 \sqrt{2}-7) \\ & \Rightarrow \alpha=162 \times 5, \beta=162 \times 7 \\ & \Rightarrow \alpha+\beta=162 \times 12=1944 \end{aligned}$
Explanation:
$\begin{aligned} & \text { Given } \frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1 \\ & \begin{aligned} & \Rightarrow b^2 x^2-a^2 y^2-12 x b^2+4 y a^2+36 b^2-4 a^2-a^2 b^2=0 \\ & \text { Comparing } \frac{b^2}{a^2}=3 \Rightarrow e^2=1+\frac{b^2}{a^2} \\ & \Rightarrow e^2=4 \\ & \Rightarrow e=2 \end{aligned} \end{aligned}$
$\begin{aligned} &\text { Similarly, } 2 a e=4\\ &\begin{aligned} \Rightarrow & a=1 \Rightarrow b=\sqrt{3} \\ & \frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1 \\ \Rightarrow & 3 x^2-y^2-36 x+4 y+108-4-3=0 \\ \Rightarrow & 3 x^2-y^2-36 x+4 y+101=0 \\ \Rightarrow & \alpha=36, \beta=4, \gamma=101 \\ \Rightarrow & \alpha+\beta+\gamma=141 \end{aligned} \end{aligned}$
Let the product of the focal distances of the point $\mathbf{P}(4,2 \sqrt{3})$ on the hyperbola $\mathrm{H}: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be 32 . Let the length of the conjugate axis of H be $p$ and the length of its latus rectum be $q$. Then $p^2+q^2$ is equal to__________
Explanation:
To find $ p^2 + q^2 $ for the hyperbola $\mathrm{H}: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, given that the product of the focal distances from a point $ \mathbf{P}(4, 2\sqrt{3}) $ to the foci is 32, we follow these steps:
Verify the Point on the Hyperbola:
The point $ P(4, 2\sqrt{3}) $ satisfies the hyperbola equation:
$ \frac{16}{a^2} - \frac{12}{b^2} = 1 \quad \ldots \text{(i)} $
Product of Focal Distances:
The distances from $ P $ to the foci $ S $ and $ S' $ are:
$ SP = e\left(4 - \frac{a}{e}\right), \quad S'P = e\left(4 + \frac{a}{e}\right) $
Therefore, the product:
$ SP \cdot S'P = 16e^2 - a^2 = 32 $
Relating $ e $, $ a $, and $ b $:
Using the identity for eccentricity:
$ e^2 = 1 + \frac{b^2}{a^2} $
Substituting in the product equation gives:
$ 16\left(1 + \frac{b^2}{a^2}\right) - a^2 = 32 \quad \ldots \text{(ii)} $
Solving for $ a^2 $ and $ b^2 $:
From equations (i) and (ii), solve for $ a^2 $ and $ b^2 $:
$ a^2 = 8, \quad b^2 = 12 $
Calculate $ p^2 + q^2 $:
The conjugate axis length is $ p = 2b $ and the latus rectum is $ q = \frac{2b^2}{a} $. Thus:
$ p^2 + q^2 = (2b)^2 + \left(\frac{2b^2}{a}\right)^2 $
Substituting $ b^2 = 12 $ and $ a^2 = 8 $ gives:
$ p^2 + q^2 = 4b^2 + \frac{4b^4}{a^2} = 48 + \frac{4 \times 144}{8} $
$ p^2 + q^2 = 48 \cdot \left(1 + \frac{12}{8}\right) = 120 $
Therefore, $ p^2 + q^2 = 120 $.
Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their ecentricities be $e_1=\sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$, then $25 \mathrm{e}_2^2$ is equal to _________ .
Explanation:
$\begin{aligned} & \frac{2 b^2}{\mathrm{a}}=15 \sqrt{2} \\ & 1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{5}{2} \\ & \mathrm{a}=5 \sqrt{2} \\ & \mathrm{~b}=5 \sqrt{3} \\ & \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=12 \sqrt{5} \\ & 2 \mathrm{a} \cdot 2 \mathrm{~B}=100 \sqrt{10} \\ & 2.5 \sqrt{2} .2 \mathrm{~B}=100 \sqrt{10} \\ & \mathrm{~B}=5 \sqrt{5} \\ & \mathrm{~A}=5 \sqrt{6} \\ & \mathrm{e}_2^2=1+\frac{\mathrm{A}^2}{\mathrm{~B}^2} \\ & =1+\frac{150}{125} \\ & \mathrm{e}_2^2=1+\frac{30}{25} \\ & 25 \mathrm{e}_2^2=55 \end{aligned}$
Let $\mathrm{S}$ be the focus of the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$, on the positive $x$-axis. Let $\mathrm{C}$ be the circle with its centre at $\mathrm{A}(\sqrt{6}, \sqrt{5})$ and passing through the point $\mathrm{S}$. If $\mathrm{O}$ is the origin and $\mathrm{SAB}$ is a diameter of $\mathrm{C}$, then the square of the area of the triangle OSB is equal to __________.
Explanation:
$\begin{aligned} & \frac{x^2}{3}-\frac{y^2}{5}=1 \\ & 5=3\left(e^2-1\right) \Rightarrow e=\sqrt{\frac{8}{3}} \\ & S \equiv(2 \sqrt{2}, 0) \end{aligned}$
$A$ is mid-point of $B S$
$\Rightarrow \quad B(2 \sqrt{6}-2 \sqrt{2}, 2 \sqrt{5})$
$\Delta (OSB) = \left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {2\sqrt 2 } & 0 & 1 \cr {2\sqrt 6 - 2\sqrt 2 } & {2\sqrt 5 } & 1 \cr } } \right|} \right| = 2\sqrt {10} $<./p>
$(\Delta(O S B))^2=40$
The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $x= \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y-\sqrt{3} x+\sqrt{3}=0$ touch this hyperbola at $\left(x_0, y_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point $\left(x_0, y_0\right)$, then $4 \mathrm{e}^2+\mathrm{m}$ is equal to _________.
Explanation:
$y=m x \pm \sqrt{a^2 m^2-b^2}$ is the tangent
$\begin{aligned} & \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\ & \Rightarrow 3 a^2-b^2=3 \end{aligned}$
$\begin{aligned} & \frac{2 b^2}{a}=9 \Rightarrow b^2=\frac{9 a}{2} \Rightarrow 3 a^2-\frac{9 a}{2}=3 \\ & \Rightarrow a^2-\frac{3}{2} a-1=0 \\ & \Rightarrow a=2 \text { or }-0.5 \text { (ignore) } \\ & \Rightarrow b=3 \\ & \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1 \\ & \Rightarrow \text { Solving hyperbola and tangent } y=\sqrt{3 x}-\sqrt{3} \\ & x_0=4, y_0=3 \sqrt{3}, e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \\ & P F_1 \cdot P F_2= \\ & \sqrt{(4-\sqrt{13})^2+(3 \sqrt{3})^2} \sqrt{(4+\sqrt{13})^2+(3 \sqrt{3})^2} \\ & =\sqrt{(56-8 \sqrt{13})(56+8 \sqrt{13})} \\ & =\sqrt{2304}=48=m \\ & \Rightarrow 4 e^2+m=13+48=61 \end{aligned}$
Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1$ equals
Explanation:
$\begin{aligned} & \text { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} \\ & a=5 \quad b^2=\frac{5 \sqrt{2} a}{2} \\ & b^2=a^2\left(1-e^2\right)=\frac{5 \sqrt{2} a}{2} \end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{a}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\ & \Rightarrow \frac{5}{\mathrm{e}}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\ & \Rightarrow \sqrt{2}-\sqrt{2} \mathrm{e}^2=\mathrm{e} \\ & \Rightarrow \sqrt{2} \mathrm{e}^2+\mathrm{e}-\sqrt{2}=0 \\ & \Rightarrow \sqrt{2} \mathrm{e}^2+2 \mathrm{e}-\mathrm{e}-\sqrt{2}=0 \\ & \Rightarrow \sqrt{2} \mathrm{e}(\mathrm{e}+\sqrt{2})-1(1+\sqrt{2})=0 \\ & \Rightarrow(\mathrm{e}+\sqrt{2})(\sqrt{2} \mathrm{e}-1)=0 \\ & \therefore \mathrm{e} \neq-\sqrt{2} ; \mathrm{e}=\frac{1}{\sqrt{2}} \end{aligned}$
$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1 \quad \mathrm{a}=5 \sqrt{2} \\ & b=5 \\ & a^2 b^2=b^2\left(e_1^2-1\right) \Rightarrow e_1^2=51 \\ & \end{aligned}$
Let the latus rectum of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $\mathrm{b}^2$ is equal to $\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$, where $l$ and $\mathrm{m}$ are co-prime numbers, then $\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2$ is equal to ________.
Explanation:
LR subtends $60^{\circ}$ at centre
$\begin{aligned} & \Rightarrow \tan 30^{\circ}=\frac{\mathrm{b}^2 / \mathrm{a}}{\mathrm{ae}}=\frac{\mathrm{b}^2}{\mathrm{a}^2 \mathrm{e}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^2}{9} \end{aligned}$
Also, $\mathrm{e}^2=1+\frac{\mathrm{b}^2}{9} \Rightarrow 1+\frac{\mathrm{b}^2}{9}=\frac{3 \mathrm{~b}^4}{81}$
$\begin{aligned} & \Rightarrow \mathrm{b}^4=3 \mathrm{~b}^2+27 \\ & \Rightarrow \mathrm{b}^4-3 \mathrm{~b}^2-27=0 \\ & \Rightarrow \mathrm{b}^2=\frac{3}{2}(1+\sqrt{13}) \\ & \Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13 \\ & \Rightarrow \ell^2+\mathrm{m}^2+\mathrm{n}^2=182 \end{aligned}$
The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$. A tangent, perpendicular to the line $2 x+3 y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $\mathrm{x}$ - and $\mathrm{y}$-axes are $\mathrm{a}$ and $\mathrm{b}$ respectively, then $|6 a|+|5 b|$ is equal to __________
Explanation:
Given that the foci are at $(\pm 2, 0)$, the distance between the foci is $2ae = 2\times2 = 4$.
So, $a = \frac{4}{3}$.
The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e = \frac{3}{2}$.
For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$, which gives $b^2 = a^2(e^2 - 1)$.
Substituting the given values, we get $b^2 = \frac{16}{9} - \frac{16}{9} = \frac{20}{9}$.
The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, or $\frac{9x^2}{16} - \frac{9y^2}{20} = 1$.
The slope of the tangent line, which is perpendicular to the given line $2x + 3y = 6$, is $\frac{3}{2}$.
The equation of the tangent to a hyperbola is $y = mx \pm \sqrt{a^2 m^2 - b^2}$. Substituting the given values, we get $y = \frac{3x}{2} \pm \sqrt{\frac{16}{9}\times\frac{9}{4} - \frac{20}{9}} = \frac{3x}{2} \pm \frac{4}{3}$.
The $x$-intercept occurs when $y = 0$, so $|a| = \frac{8}{9}$, and the $y$-intercept occurs when $x = 0$, so $|b| = \frac{4}{3}$.
Finally, $|6a| + |5b| = 6\frac{8}{9} + 5\frac{4}{3} = \frac{48}{9} + \frac{60}{9} = \frac{108}{9} = 12$.
Let $m_{1}$ and $m_{2}$ be the slopes of the tangents drawn from the point $\mathrm{P}(4,1)$ to the hyperbola $H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1$. If $\mathrm{Q}$ is the point from which the tangents drawn to $\mathrm{H}$ have slopes $\left|m_{1}\right|$ and $\left|m_{2}\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis, then $\frac{(P Q)^{2}}{\alpha \beta}$ is equal to __________.
Explanation:
$ y=m x \pm \sqrt{a^2-b^2 m^2} $
Given the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as :
$y=mx \pm \sqrt{25 - 16m^2}$
We know that the tangents pass through the point $P(4, 1)$, which gives us the equation :
$1 = 4m \pm \sqrt{25 - 16m^2}$
Squaring both sides to get rid of the square root, we obtain :
$(4m - 1)^2 = 25 - 16m^2$
which simplifies to the quadratic equation :
$4m^2 - m - 3 = 0.$
Solving this equation, we find the roots $m_1 = 1$ and $m_2 = -\frac{3}{4}$, which are the slopes of the tangents.
Given that we are interested in the positive values of the slopes, we consider $|m_1| = 1$ and $|m_2| = \frac{3}{4}$.
The equations of the tangents are then :
1) $y = x - 3$ and
2) $y = \frac{3}{4}x - 4$.
The x-intercepts of these lines (when $y = 0$) are given by $x = \alpha = 3$ for the first line and $x = \beta = \frac{16}{3}$ for the second line.
The intersection point $Q$ of these tangents is found by solving the system of equations $y = x - 3$ and $y = \frac{3}{4}x - 4$, which gives $Q(-4, -7)$.
The square of the distance $PQ$ is then $(-4-4)^2 + (-7-1)^2 = 128$.
Therefore, $\frac{(PQ)^2}{\alpha \beta} = \frac{128}{3 \times \frac{16}{3}} = 8$
Let $\mathrm{H}_{\mathrm{n}}: \frac{x^{2}}{1+n}-\frac{y^{2}}{3+n}=1, n \in N$. Let $\mathrm{k}$ be the smallest even value of $\mathrm{n}$ such that the eccentricity of $\mathrm{H}_{\mathrm{k}}$ is a rational number. If $l$ is the length of the latus rectum of $\mathrm{H}_{\mathrm{k}}$, then $21 l$ is equal to ____________.
Explanation:
$ H_n \Rightarrow \frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n \in N $
Here, $a^2=1+n$ and $b^2=3+n$
$ \begin{aligned} \operatorname{Eccentricity}(e) & =\sqrt{1+\frac{b^2}{a^2}} \\\\ & =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}} \end{aligned} $
The smallest even value for which $e \in Q$ is 48 .
$ \begin{aligned} \therefore n & =48 \\\\ \therefore e & =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7} \end{aligned} $
$ \begin{array}{ll} \Rightarrow a^2=n+1=49, b^2=n+3=51 \\\\ \therefore 21 l=21 \times\left(\frac{2 b^2}{a}\right)=21 \times 2 \times \frac{51}{7}=306 \end{array} $
Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is reciprocal to that of the hyperbola $2 x^{2}-2 y^{2}=1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ___________.
Explanation:
$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
Here, $a=b$
$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$
$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$
Since, ellipse intersects the hyperbola at right angles
$\therefore$ Ellipse and the hyperbola are confocal
$\therefore$ Foci of hyperbola $=(1,0)$
$\Rightarrow$ Foci of ellipse, $a e=1$
$ \begin{array}{ll} &\Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1 \\\\ &\Rightarrow a=\sqrt{2} \\\\ &\therefore e^2=1-\frac{b^2}{a^2} \\\\ &\Rightarrow \frac{1}{2}=1-\frac{b^2}{2} \\\\ &\Rightarrow \frac{b^2}{2}=\frac{1}{2} \\\\ &\Rightarrow b^2=1 \end{array} $
$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\therefore$ Square of length of the latus rectum $=2$
The vertices of a hyperbola H are ($\pm$ 6, 0) and its eccentricity is ${{\sqrt 5 } \over 2}$. Let N be the normal to H at a point in the first quadrant and parallel to the line $\sqrt 2 x + y = 2\sqrt 2 $. If d is the length of the line segment of N between H and the y-axis then d$^2$ is equal to _____________.
Explanation:
$ H: \frac{x^2}{36}-\frac{y^2}{9}=1 $
equation of normal is $6 x \cos \theta+3 y \cot \theta=45$
$ \begin{aligned} & \text { slope }=-2 \sin \theta=-\sqrt{2} \\\\ & \Rightarrow \theta=\frac{\pi}{4} \end{aligned} $
Equation of normal is $\sqrt{2} x+y=15$
$P:(a \sec \theta, b \tan \theta)$
$\Rightarrow \mathrm{P}(6 \sqrt{2}, 3)$ and $\mathrm{K}(0,15)$
$ d^2=216 $
For the hyperbola $\mathrm{H}: x^{2}-y^{2}=1$ and the ellipse $\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$, a $>\mathrm{b}>0$, let the
(1) eccentricity of $\mathrm{E}$ be reciprocal of the eccentricity of $\mathrm{H}$, and
(2) the line $y=\sqrt{\frac{5}{2}} x+\mathrm{K}$ be a common tangent of $\mathrm{E}$ and $\mathrm{H}$.
Then $4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$ is equal to _____________.
Explanation:
The equation of tangent to hyperbola ${x^2} - {y^2} = 1$ within slope $m$ is equal to $y = mx\, \pm \,\sqrt {{m^2} - 1} $ ...... (i)
And for same slope $m$, equation of tangent to ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ is $y = mx\, \pm \,\sqrt {{a^2}{m^2} + {b^2}} $ ...... (ii)
$\because$ Equation (i) and (ii) are identical
$\therefore$ ${a^2}{m^2} + {b^2} = {m^2} - 1$
$\therefore$ ${m^2} = {{1 + {b^2}} \over {1 - {a^2}}}$
But equation of common tangent is $y = \sqrt {{5 \over 2}} x + k$
$\therefore$ $m = \sqrt {{5 \over 2}} \Rightarrow {5 \over 2} = {{1 + {b^2}} \over {1 - {a^2}}}$
$\therefore$ $5{a^2} + 2{b^2} = 3$ ....... (i)
eccentricity of ellipse $ = {1 \over {\sqrt 2 }}$
$\therefore$ $1 - {{{b^2}} \over {{a^2}}} = {1 \over 2}$
$ \Rightarrow {a^2} = 2{b^2}$ ....... (ii)
From equation (i) and (ii) : ${a^2} = {1 \over 2},\,{b^2} = {1 \over 4}$
$\therefore$ $4({a^2} + {b^2}) = 3$
A common tangent $\mathrm{T}$ to the curves $\mathrm{C}_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $\mathrm{T}$ touches $\mathrm{C}_{1}$ at $\left(x_{1}, y_{1}\right)$ and $\mathrm{C}_{2}$ at $\left(x_{2}, y_{2}\right)$, then $\left|2 x_{1}+x_{2}\right|$ is equal to ______________.
Explanation:
Equation of tangent to ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ and given slope m is : $y = mx + \sqrt {4{m^2} + 9} $ ..... (i)
For slope m equation of tangent to hyperbola is :
$y = mx + \sqrt {42{m^2} - 143} $ ....... (ii)
Tangents from (i) and (ii) are identical then
$4{m^2} + 9 = 42{m^2} - 143$
$\therefore$ $m = \, \pm \,2$ (+2 is not acceptable)
$\therefore$ $m = - 2$.
Hence, ${x_1} = {8 \over 5}$ and ${x_2} = {{84} \over 5}$
$\therefore$ $|2{x_1} + {x_2}| = \left| {{{16} \over 5} + {{84} \over 5}} \right| = 20$
An ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the vertices of the hyperbola $H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$. Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$, respectively. Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$. If $l$ is the length of the latus rectum of the ellipse $E$, then the value of $113 l$ is equal to _____________.
Explanation:
Vertices of hyperbola $ = (0,\, \pm \,8)$
As ellipse pass through it i.e.,
$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$ ...... (1)
As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.
${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}$
and ${e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}$
$\therefore$ ${e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}$
$ \Rightarrow (64 - {a^2})(113) = {32^2}$
$ \Rightarrow {a^2} = 64 - {{1024} \over {113}}$
L.R of ellipse $ = {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)$
$ = l = {{1552} \over {113}}$
$\therefore$ $113l = 1552$
Let the equation of two diameters of a circle $x^{2}+y^{2}-2 x+2 f y+1=0$ be $2 p x-y=1$ and $2 x+p y=4 p$. Then the slope m $ \in $ $(0, \infty)$ of the tangent to the hyperbola $3 x^{2}-y^{2}=3$ passing through the centre of the circle is equal to _______________.
Explanation:
Hyperbola $3 x^2-y^2=3, x^2-\frac{y^2}{3}=1$
$ \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{m}^2-3} $
It passes $(1,0)$
$o=m \pm \sqrt{m^2-3}$
$m$ tends $\infty$
$ \begin{aligned} &\text {It passes }(1,3) \\\\ &3=m \pm \sqrt{m^2-3} \\\\ &(3-m)^2=m^2-3 \\\\ &m=2 \end{aligned} $
Let $H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2\sqrt 2 + \sqrt {14} )$. If the eccentricity H is ${{\sqrt {11} } \over 2}$, then the value of a2 + b2 is equal to __________.
Explanation:
$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$ ...... (1)
$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$ ....... (2)
$ \Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$ ....... (3)
and $a + b = 4\sqrt 2 + 2\sqrt {14} $ ...... (4)
By (3) and (4)
$ \Rightarrow a + {{\sqrt 7 } \over 2}a = 4\sqrt 2 + 2\sqrt {14} $
$ \Rightarrow {{a\left( {2 + \sqrt 7 } \right)} \over 2} = 2\sqrt 2 \left( {2 + \sqrt 7 } \right)$
$ \Rightarrow a = 4\sqrt 2 \Rightarrow {a^2} = 32$ and ${b^2} = 56$
$ \Rightarrow {a^2} + {b^2} = 32 + 56 = 88$
Let a line L1 be tangent to the hyperbola ${{{x^2}} \over {16}} - {{{y^2}} \over 4} = 1$ and let L2 be the line passing through the origin and perpendicular to L1. If the locus of the point of intersection of L1 and L2 is ${({x^2} + {y^2})^2} = \alpha {x^2} + \beta {y^2}$, then $\alpha$ + $\beta$ is equal to _____________.
Explanation:
Equation of L1 is
${{x\sec \theta } \over 4} - {{y\tan \theta } \over 2} = 1$ ..... (i)
Equation of line L2 is
${{x\tan \theta } \over 2} + {{y\sec \theta } \over 4} = 0$ ..... (ii)
$\because$ Required point of intersection of L1 and L2 is (x1, y1) then
${{{x_1}\sec \theta } \over 4} - {{{y_1}\tan \theta } \over 2} - 1 = 0$ ...... (iii)
and ${{{y_1}\sec \theta } \over 4} - {{{x_1}\tan \theta } \over 2} = 0$ ...... (iv)
From equations (iii) and (iv)
$\sec \theta = {{4{x_1}} \over {x_1^2 + y_1^2}}$ and $\tan \theta = {{ - 2{y_1}} \over {x_1^2 + y_1^2}}$
$\therefore$ Required locus of (x1, y1) is
${({x^2} + {y^2})^2} = 16{x^2} - 4{y^2}$
$\therefore$ $\alpha$ = 16, $\beta$ = $-$4
$\therefore$ $\alpha$ + $\beta$ = 12
Let the eccentricity of the hyperbola ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ be ${5 \over 4}$. If the equation of the normal at the point $\left( {{8 \over {\sqrt {5} }},{{12} \over {5}}} \right)$ on the hyperbola is $8\sqrt 5 x + \beta y = \lambda $, then $\lambda$ $-$ $\beta$ is equal to ___________.
Explanation:
${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\left( {e = {5 \over 4}} \right)$
So, ${b^2} = {a^2}\left( {{{25} \over {16}} - 1} \right) \Rightarrow b = {3 \over 4}a$
Also $\left( {{8 \over {\sqrt 5 }},{{12} \over 5}} \right)$ lies on the given hyperbola
So, ${{64} \over {5{a^2}}} - {{144} \over {25\left( {{{9{a^2}} \over {16}}} \right)}} = 1 \Rightarrow a = {8 \over 5}$ and $b = {6 \over 5}$
Equation of normal
${{64} \over {25}}\left( {{x \over {{8 \over {\sqrt 5 }}}}} \right) + {{36} \over {25}}\left( {{y \over {{{12} \over 5}}}} \right) = 4$
$ \Rightarrow {8 \over {5\sqrt 5 }}x + {3 \over 5}y = 4$
$ \Rightarrow 8\sqrt 5 x + 15y = 100$
So, $\beta$ = 15 and $\lambda$ = 100
Gives $\lambda$ $-$ $\beta$ = 85
Let the hyperbola $H:{{{x^2}} \over {{a^2}}} - {y^2} = 1$ and the ellipse $E:3{x^2} + 4{y^2} = 12$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If ${e_H}$ and ${e_E}$ are the eccentricities of H and E respectively, then the value of $12\left( {e_H^2 + e_E^2} \right)$ is equal to ___________.
Explanation:
$\because$ $H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 1} = 1$
$\therefore$ Length of latus rectum $ = {2 \over a}$
$E:{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$
Length of latus rectum $ = {6 \over 2} = 3$
$\because$ ${2 \over a} = 3 \Rightarrow a = {2 \over 3}$
$\therefore$ $12\left( {e_H^2 + e_E^2} \right) = 12\left( {1 + {9 \over 4}} \right) + \left( {1 - {3 \over 4}} \right) = 42$
Explanation:
Therefore, 2sec2$\theta$ $-$ 4tan2$\theta$ = 2
$\Rightarrow$ 2 + 2tan2$\theta$ $-$ 4tan2$\theta$ = 2
$\Rightarrow$ tan$\theta$ = 0 $\Rightarrow$ $\theta$ = 0
Similarly, for point B, we will get $\phi$ = 0.
but according to question $\theta$ + $\phi$ = ${\pi \over 2}$ which is not possible.
Hence, it must be a 'BONUS'.
Explanation:
$\sqrt 3 kx - ky = 4\sqrt 3 {k^2}$ ....... (2)
Adding equation (1) & (2)
$2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)$
$x = 2\left( {k + {1 \over k}} \right)$ ......... (3)
Substracting equation (1) & (2)
$y = 2\sqrt 3 \left( {{1 \over k} - k} \right)$ ........(4)
$\therefore$ ${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 4$
${{{x^2}} \over {16}} - {{{y^2}} \over {48}} = 1$ (Hyperbola)
$ \therefore $ ${e^2} = 1 + {{48} \over {16}}$
$ \Rightarrow $ $e = 2$
Column $I$
(A) Two intersecting circles
(B) Two mutually external circles
(C) Two circles, one strictly inside the other
(D) Two branches vof a hyperbola
Column $II$
(p) have a common tangent
(q) have a common normal
(r) do not have a common tangent
(s) do not have a common normal
| List - I | List - II | ||
|---|---|---|---|
| P. | The length of the conjugate axis of H is | 1. | 8 |
| Q. | The eccentricity of H is | 2. | ${4 \over {\sqrt 3 }}$ |
| R. | The distance between the foci of H is | 3. | ${2 \over {\sqrt 3 }}$ |
| S. | The length of the latus rectum of H is | 4. | 4 |
