Conic Sections

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2sin\theta -3}{2}=h\frac{3cos\theta -5}{2}=k$

sin² θ + cos² θ = 1

${\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2$ = 1

$36x^2+16y^2+108x+80y+145=0$

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$

0 = -16 m + c . (i)

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

(i) & (ii) Þ 36 m² = 4 (m² + 1)

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

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

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