Conic Sections

0 = 16 m + c . (i)

 $\frac{\left|m\left(?10\right)+C\right|}{\sqrt[]{m^2+1}}=2.............\left(ii\right)$

(i) & (ii) 36 m2 = 4 (m2 + 1)

$m=\frac{1}{2\sqrt[]{2}},C=\frac{8}{\sqrt[]{2}}$

$4\sqrt[]{2}\left(m+c\right)=34$

Since A (sec θ, 2 tanθ) & B $\left(sec?,2tan?\right)$  lies on 2x² – y² = 2 then

2sec² θ - 4tan^2_θ = 2 or sec² θ - 2 tan² θ = 1

-> $ta{n}^2\theta =0so\theta =0$

Similarly    $?=0but\theta +?=\frac{\pi }{2}$ (given) so not possible

Hence question is not correct

Equation of tangent at P (2, -4)

y (-4) = 4 (x + 2)

x + y + 2 = 0

So, A (-2, 0)

Equation of normal at P:

y + 4 = 1 (x – 2)

x – y = 6

So, B (-2, -8)

For square mid-point of AB = mid-point of PQ

$\Rightarrow \frac{a+2}{2}=-2\Rightarrow a=-6$        

$\frac{b-4}{2}=\frac{-8}{2}\Rightarrow b=-4$   

So, 2a + b = -16

Locus of point of intersection of perpendicular tangent will be its director circle & director circle of parabola be its directrix.

Given parabola y² = 16 (x – 3) so equation of directrix be x – 3 = - 4 i.e., x + 1 = 0

(y – 2)² = (x – 1)

2 (y – 2)    $\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}\left(2,3\right)=\frac{1}{2\left(3-2\right)}=\frac{1}{2}$              

Equation of tangent at P (2, 3):

$y-3=\frac{1}{2}\left(x-2\right)$  

2y – 6 = x – 2

x – 2y + 4 = 0

Q (-4, 0)

Required area = ${\displaystyle \underset{0}{\overset{3}{\int }}\left({\left(y-2\right)}^2+1-\left(2y-4\right)\right)dy=9}$

Equation of tangent to given ellipse at

$P:\frac{x.cos\theta }{b}+\frac{y.sin\theta }{2a}=1$

A (b sec θ. 0) 7 B (0, 2a cosec θ)

area of    $\Delta OAB$

$=\frac{2ab}{2sin\theta cos\theta }=\frac{2ab}{sin2\theta }$

For minimum area sin 2θ = 1

So minimum area = 2ab

=>k = 2

$r=\sqrt[]{{\left(\frac{p}{2}\right)}^2+{\left(\frac{1-p}{2}\right)}^2-5}=\sqrt[]{\frac{p^2+1+p^2-2p-20}{4}}=\sqrt[]{\frac{2p^2-2p-19}{2}}$

Since $r\in \left(0,5\right)so0<\sqrt[]{2p^2-2p-19}<10$

$\Rightarrow 2p^2-2p-19>0\&2p^2-2p-19<100$     

$P\in \left[\frac{1-\sqrt[]{239}}{2},\frac{1-\sqrt[]{39}}{2}\right)\cup \left(\frac{1+\sqrt[]{39}}{2},\frac{1+\sqrt[]{239}}{2}\right]$

So number of integral values of P² is 61.

$P\left(-2\sqrt[]{6},\sqrt[]{3}\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$              

$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1.........\left(i\right)$              

$b^2=\frac{a^2}{4}.........\left(ii\right)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$  

Cuts conjugate axis at $R\left(0,R\sqrt[]{3}\right)$     

$\therefore QR=6\sqrt[]{3}$        

Equation of required circle

$C:{\left(x-2\right)}^2+{\left(y-1\right)}^2+\lambda \left(2y-x\right)=0..........\left(i\right)$              

Intersect C₁x² + y² + 2y – 5 = 0 .(ii)

$\therefore$ Equation of radical axis

    $-4x-4y+10+\lambda \left(2y-x\right)=0..........\left(ii\right)$          

Centre of C₁ (0, -1) lies on .(ii)

$\Rightarrow 4+10-2\lambda =0\Rightarrow \lambda =7$        

Equation of circle C is

$x^2+y^2-11x+12y+5=0$       

Diameter = $\sqrt[]{245}=7\sqrt[]{5}$