Conic Sections

$y^{2}dx+\left(x^2?xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2?xy+y^2}{y^2}=0?\frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2?\left(\frac{x}{y}\right)+1=0$

Put x = vy $?v+y\frac{dv}{dy}+v^2?v+1=0$

$?\frac{ydv}{dy}+v^2+1=0$

$?\frac{?}{6}+\mathcal{l}n\left|y\right|=\frac{?}{4}$

$\mathcal{l}n\left|y\right|=\frac{?}{12}$

$y^{2}dx+\left(x^2?xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2?xy+y^2}{y^2}=0?\frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2?\left(\frac{x}{y}\right)+1=0$

Put x = vy $?v+y\frac{dv}{dy}+v^2?v+1=0$ $\frac{}{}{}^{}$

$?\frac{ydv}{dy}+v^2+1=0$

$?\frac{?}{6}+\mathcal{l}n\left|y\right|=\frac{?}{4}$

$\mathcal{l}n\left|y\right|=\frac{?}{12}$

$y^{2}dx+\left(x^2?xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2?xy+y^2}{y^2}=0?\frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2?\left(\frac{x}{y}\right)+1=0$

Put x = vy $?v+y\frac{dv}{dy}+v^2?v+1=0$ $\frac{}{}{}^{}$

$?\frac{ydv}{dy}+v^2+1=0$

$?\frac{?}{6}+\mathcal{l}n\left|y\right|=\frac{?}{4}$

$\mathcal{l}n\left|y\right|=\frac{?}{12}$

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

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

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

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

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

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

^{$x-ty+\frac{1}{4}{t}^2=0$}

^{$\frac{1}{t}=m,m_2\Rightarrow \left(8m_1{m}_2\right)=3\sqrt[]{2}-4$}

$\frac{x^2}{\frac{1}{25}}+\frac{y^2}{\frac{1}{4}}=1$      

$25{\alpha }^2+4{\beta }^2=1..........(i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$  passes

though (a, b)

$\alpha m^2-\beta m+1=0$

$m_1+m_2=\frac{\beta }{\alpha },4m_1^2=\frac{1}{\alpha }$

$\therefore from(i)\&(ii)$

25(a² +  a) = 1 …………(iii)

$\therefore {\left(10\alpha +5\right)}^2+{\left(16{\beta }^2+50\right)}^2=2929$

$\therefore a=1+0+b{\left(sin-\frac{\pi }{2}\right)}^{\frac{\pi }{2}}+2b+1$           

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

S (4, 4) and V (3, 2)

$\therefore$ point of intersection of directrix with axis of parabola is A (2, 0)

Image of A (2, 0) with respect to line

$\therefore x+2y=6isB\left(x_2,y_2\right)$

$\therefore \frac{x_2-2}{1}=\frac{y_2-0}{2}\Rightarrow \frac{-2\left(2+0-6\right)}{5}$

$\therefore B\left(\frac{18}{5},\frac{16}{5}\right)$

Point B is point of intersection of directrix with axes of parabola P₂.

$\therefore x+2y=\lambda$

$B\left(\frac{18}{5},\frac{16}{5}\right)$ lies on the line x + 2y = $\lambda$  $\therefore \lambda =\frac{18}{5}+\frac{32}{5}=10$