Conic Sections

C : (x – 2)² + y² = 1

Equation of chord AB : 2x = 3

$OA=OB=\sqrt[]{3}$

$AM=\frac{\sqrt[]{3}}{2}$

$Areaof\Delta OAB=\frac{1}{2}\left(2AM\right)\left(OM\right)$

$=\frac{3\sqrt[]{3}}{4}sq.unit$

$a^2\left(e^2-1\right)=b^2$  

$e=\sqrt[]{\frac{5}{2}}\Rightarrow b^2=\frac{3a^2}{2}$                

Length of latus rectum $\frac{2b^2}{a}=6\sqrt[]{2}$  

$3a=6\sqrt[]{2}\Rightarrow a=2\sqrt[]{2}$                

$b=2\sqrt[]{3}$                

y = 2x + c is tangent to hyperbola

$\therefore c^2=a^2{m}^2-b^2=20$        

Given vertex is  (5, 4) and directrix 3x + y – 29 = 0

Let foot of perpendicular of (5, 4) on directrix be (x₁, y₁)

$\frac{x_1-5}{3}=\frac{y_1-4}{1}=\frac{-\left(-10\right)}{10}$               

$\therefore \left(x_1,y_1\right)=\left(8,5\right)$               

So, focus of parabola will be S = (2, 3)

Let P(x, y) be any point on parabola, then

$\Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2=\frac{{\left(3x+y-29\right)}^2}{10}$               

$\Rightarrow x^2+9y^2-6xy+134x-2y-711=0$               

And given parabola

$x^2+a{y}^2+bxy+cx+dy+k=0$               

$\therefore a=9,b=-6,c=134,d=-2,k=-711$               

$\therefore a+b+c+d+k=576$    

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

^{$\begin{array}{l}Giventhate=\frac{3}{2}\\ andfoci=\left(\pm ae,0\right)=\left(\pm 2,0\right)\\ \therefore ae=2\Rightarrow a*\frac{3}{2}=2\Rightarrow a=\frac{4}{3}\\ Weknowthat{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=\frac{16}{9}\left(\frac{9}{4}-1\right)\Rightarrow b^2=\frac{16}{9}*\frac{5}{4}\\ \Rightarrow b^2=\frac{20}{9}\\ So,theequationofthehyperbolais\\ \frac{x^2}{{\left(4/3\right)}^2}-\frac{y^2}{20/9}=1\Rightarrow \frac{9x^2}{16}-\frac{9y^2}{20}=1\Rightarrow \frac{x^2}{16}-\frac{y^2}{20}=\frac{1}{9}\\ \Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$}

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$Weknowthat\text{\hspace{0.17em}Distance }$
$\begin{array}{l}betweenthefoci=2ae\\ \therefore 2ae=16\Rightarrow ae=8\\ Giventhate=\sqrt[]{2}\\ \therefore \sqrt[]{2}a=8\Rightarrow a=4\sqrt[]{2}\\ Now{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=32\left(2-1\right)\Rightarrow b^2=32\\ So,theequationofthehyperbolais\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow \frac{x^2}{32}-\frac{y^2}{32}=1\Rightarrow x^2-y^2=32\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lengthofthelatusrectumofthehyperbola\\ =\frac{2b^2}{a}=8\Rightarrow b^2=4a\dots \left(i\right)\\ \text{\hspace{0.17em}Distance}betweenthefoci=2ae\\ Transverseaxis=2a\\ andConjugateaxis=2b\\ \therefore \frac{1}{2}\left(2ae\right)=2b\Rightarrow ae=2b\Rightarrow b=\frac{ae}{2}\dots \left(ii\right)\\ \Rightarrow b^2=\frac{a^2{e}^2}{4}\Rightarrow 4a=\frac{a^2{e}^2}{4}\left[Fromeqn.\left(i\right)\right]\\ \Rightarrow 16=a{e}^2\therefore a=\frac{16}{e^2}\\ Now{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow 4a=a^2\left(e^2-1\right)\\ \Rightarrow \frac{4}{a}=e^2-1\Rightarrow \frac{4}{16/e^2}=e^2-1\\ \Rightarrow \frac{e^2}{4}=e^2-1\Rightarrow e^2-\frac{e^2}{4}=1\Rightarrow \frac{3e^2}{4}=1\\ \Rightarrow e^2=\frac{4}{3}\therefore e=\frac{2}{\sqrt[]{3}}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\left(a<b\right)\\ \therefore Eccentricitye=\sqrt[]{1-\frac{a^2}{b^2}}\Rightarrow e^2=1-\frac{a^2}{b^2}\\ \Rightarrow \frac{a^2}{b^2}=\left(1-e^2\right)\Rightarrow a^2=b^2\left(1-e^2\right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$