Conic Sections

$P\left(\frac{15}{2}{t}^2,15t\right)$

C (-15,0)

Q (-30,0)

tanθ = $\frac{PN}{QN}$

=> $\frac{1}{t}=\frac{15t}{30+\frac{15}{2}{t}^2}$

=> t = 2

$Tangent=y=\frac{1}{2}\left(x+30\right)$

$T_{\left(P\right)}:\frac{x{x}_1}{8}+\frac{y{y}_1}{4}=1,$

where $\frac{x_{1}2}{8}+\frac{y_{1}2}{4}=1...........\left(i\right)$

given : slope = 2

=> x₁ = -4y         . (ii)

(i) & (ii) => P $\left(-\frac{8}{3},\frac{2}{3}\right)$

$e=\frac{1}{\sqrt[]{2}}$         

A = ar (PSS') = $\frac{4}{3}$

$\left(5-e^2\right)A=6$

$x^2+y^2+{\left(x-1\right)}^2+y^2+x^2+{\left(y-1\right)}^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=18$

$\Rightarrow x^2+y^2-x-y-\frac{7}{2}=0$

$r^2=\frac{1}{4}+\frac{1}{4}+\frac{7}{2}=4$

Let P (h, k) be the mid point of the chord x² – y² = 4

$\therefore$ its equation is xh – yk = h² – k²

$Or,y=\left(\frac{h}{x}\right)x+\frac{k^2-h^2}{k}$ if this line is tangent to y² = 8x then $\frac{k^2-h^2}{k}$ = $\frac{2}{h/k}=\frac{2k}{h}$

$h\left(k^2-h^2\right)=2k^2$              

$\therefore$ Required locus is 2y² = x (y² – x²)

$\Rightarrow x^3=y^2\left(x-2\right)$

x = 2t $A\left(2t,\frac{t^2}{3}\right)$ $\frac{{}^{}}{}$

$\frac{{}^{}}{}$ $y=\frac{t^2}{3}$  S (0, 3)

${\frac{}{}}^{}$  $3y={\left(\frac{x}{2}\right)}^2$ $B\left(0,\lambda \right)$ $$

$3k=\frac{t^2}{3}+3+\lambda =\frac{2t^2}{3}+3-\frac{12t^2}{9-t^2}$

$\underset{t\to 1}{lim}3k=\frac{2}{3}+3-\frac{12}{8}=3-\frac{5}{6}=\frac{13}{6}$

$e=\frac{5}{4}$

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow b=\frac{3a}{4}$

$p\left(\frac{8}{\sqrt[]{5}},\frac{12}{5}\right)$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1$

$\Rightarrow \frac{\sqrt[]{5}x}{3}+\frac{5}{8}y=\frac{8}{3}+\frac{3}{2}=\frac{25}{6}$

  $\frac{x^2}{4}+\frac{y^2}{2}=1$              

lim PQ is y = x + 1

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

$\Rightarrow {\left(x_1-x_2\right)}^2={\left(-\frac{4}{3}\right)}^2-4\left(-\frac{2}{3}\right)=\frac{16}{9}+\frac{8}{3}=\frac{40}{9}$               

y = x + 1

$y_1-y_2=x_1-x_2$

$P{Q}^2=\frac{80}{9}\Rightarrow \frac{PQ}{2}=r=\frac{\sqrt[]{80}}{6}$

$r=\frac{4\sqrt[]{5}}{6}\Rightarrow {\left(3r\right)}^2={\left(2\sqrt[]{5}\right)}^2=20$               

y = x – x²

p (t, t - t²)

$T_{\left(p\right)}:y+t-t^2$

$=x+t-2x.t$

$\Rightarrow y=\left(1-2t\right)x+t^2$

$\Rightarrow 1-2t=k,t^2=4,t=\pm 2$

$\Rightarrow y=5x+4$                

$x=\frac{1}{2}$

$y=2{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(\frac{y^2+3}{4}-\frac{y^2+1}{2}\right)dy}$

$=\frac{1}{2}{\left(\frac{y^3}{3}+3y\right)}_0^1-{\left(\frac{y^3}{3}+y\right)}_0^1$

$=\frac{1}{2}\left(\frac{10}{3}\right)-\frac{4}{3}=\frac{1}{3}$

Let inclination of required line is,

So the coordinates of point B can be assumed as

$\left(4-\frac{\sqrt[]{29}}{3}cos\theta ,3-\frac{\sqrt[]{29}}{3}sin\theta \right)$

which satisfies x – y – 2 = 0

$sin\theta -cos\theta =\frac{3}{\sqrt[]{29}}$

By squaring

$sin2\theta =\frac{20}{29}=\frac{2tan\theta }{1+ta{n}^2\theta }$

Point B : $\left(\frac{10}{3},\frac{4}{3}\right)$ $\frac{}{}\frac{}{}$

which also satisfies x + 2y = 6