Conic Sections

Ellipse : $\frac{x^2}{16}+\frac{y^2}{7}=1$

Eccentricity = $\sqrt[]{1-\frac{7}{10}}=\frac{3}{4}$

Foci $\equiv \left(\pm ae,0\right)$ $\equiv \left(\pm 3,0\right)$

Hyperbola : $\frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha }{25}\right)}=1$

Eccentricity = $\sqrt[]{1+\frac{\alpha }{144}}=\frac{1}{12}\sqrt[]{144+\alpha }$

$=2.\frac{\frac{81}{25}}{\frac{12}{5}}=\frac{27}{10}$

$y^2-2x-2y=1$

$\Rightarrow {\left(y-1\right)}^2=2\left(x+1\right)$ …. (i)

Equation of tangent at A is 2x – y – 5 = 0 ………. (ii)

D is mid point of AB solving (ii) with y = 1 P (3, 1)

$\therefore PD=4,AD=2$

$Areaof\Delta APD=\frac{1}{2}\left(PD\right)\left(AD\right)=4$

$\therefore Areaof\Delta APB=8sq.units$

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the points (7, 0) & (0,  $-2\sqrt[]{6}$ )

$\therefore a^2=49\&b^2=24$

$\therefore e=\sqrt[]{1-\frac{b^2}{a^2}}=\sqrt[]{1-\frac{24}{49}}=\frac{5}{7}$

$x^2+y^2-x+2y=\frac{11}{4}$

${\left(x-\frac{1}{2}\right)}^2+{\left(y+1\right)}^2={\left(2\right)}^2$

or $\Delta PQRPR=QRsin2\ge \frac{1}{3}$

$=4.6sin\frac{\pi }{8}$

$as\Delta PQR=\frac{1}{2}PR*PQ$

$=4sin\frac{\pi }{4}=\frac{4}{\sqrt[]{2}}=2\sqrt[]{2}$

$y^2=2x-3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3              ……… (ii)

From (i) and (ii)

y = 1 or 3

$MPQ=\frac{4}{4}=1$

  $MQR=\frac{2}{6}=\frac{1}{3}$              

$M_{PQ}*M_{PR}=-1\Rightarrow PQ\perp PR$      

Orthocentre = P (2, -1)

Let C be the centre and M be the mid point of AB

$\Delta APC:sin\theta =\frac{13/2}{pc}=\frac{5}{13}\Rightarrow pC=\frac{169}{10}$                

$\Delta AMC:cos\theta =\frac{6}{13/2}=\frac{12}{13}$              

PC = $\frac{169}{10},MC=\frac{13}{2}sin\theta =\frac{13}{2}\cdot \frac{5}{13}$  

PM = PC – MC = $\frac{169}{10}-\frac{5}{2}=\frac{144}{10}$  

5PM = 72

 $\frac{{\left(x+1\right)}^2}{\lambda +5}=\frac{{\left(y+1\right)}^2}{\frac{\lambda +5}{4}}=1$

length of latus rectum = $\frac{2b^2}{a}=\frac{2\left(\frac{\lambda +5}{4}\right)}{\sqrt[]{5+\lambda }}=4$

$\sqrt[]{\lambda +5}=8\Rightarrow \lambda =59$

Major axis = $2\sqrt[]{\lambda +5}=16$

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

The circle $x^2+y^2+6x+8y+16=0$ has centre (3, 4) and radius 3 units

The circle

$x^2+y^2+2\left(3-\sqrt[]{3}\right)x+2\left(4-\sqrt[]{6}\right)y=k+6\sqrt[]{3}+8\sqrt[]{6},k>0$ has centre $\left(\sqrt[]{3}-3,\sqrt[]{6}-4\right)$ and radius $\sqrt[]{k+34}$

$?$ These two circles touch internally hence

$\sqrt[]{3+6}=\left|\sqrt[]{k+34}-3\right|$

here, k = 2 is only possible $\left(?k>0\right)$

Equation of common tangent to two circle is

$2\sqrt[]{3}x+2\sqrt[]{6}y+16+6\sqrt[]{3}+8\sqrt[]{6}+k=0$

$?k=2$ then equation is

$x+\sqrt[]{2}y+3+4\sqrt[]{2}+3\sqrt[]{3}=0$ ….(i)

$?\left(\alpha ,\beta \right)$ are foot of perpendicular from (3, 4) to line (i) then

$\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt[]{2}}=\frac{-3-4\sqrt[]{2}+3+4\sqrt[]{2}+3+\sqrt[]{3}}{1+2}$

$\therefore {\left(\alpha +\sqrt[]{3}\right)}^2+{\left(\beta +\sqrt[]{6}\right)}^2=25$

Equation of tangent to the parabola at

$P\left(\frac{8}{5},\frac{6}{5}\right)$

$75x.\frac{8}{5}=160\left(y+\frac{6}{5}\right)-192$

$\Rightarrow 120x=160y$

$\Rightarrow 3x=4y$

Equation of circle touching the given parabola at P can be taken as

$\Rightarrow \lambda =\frac{2}{5}or-\frac{8}{5}$

Radius = 1 or 4

Sum of diameter = 10