Conic Sections

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

here AB = $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area = $\frac{1}{2}*\sqrt[]{2}*\sqrt[]{2}=1$

Tangents making angle $\frac{\pi }{4}$ with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$

So, these tangents are $\perp$ . So ASB is a focal chord.

  $?\frac{dy}{dx}+e^x\left(x^2-2\right)y=\left(x^2-2x\right)\left(x^2-2\right)e^{2x}$ $\frac{}{}{}^{}{}^{}{}^{}{}^{}{}^{}$

Here, I.F.

$=e^{{\displaystyle \int e^x\left(x^2-2\right)dx}}$

$=e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  Solution of the differential equation is

$y.e^{\left(x^2-2x\right)e^x}=\left(x^2-2x-1\right)e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  y(0) = 0

$\therefore c=1$

$\therefore y\left(2\right)=-1+1=0$

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab);\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now, $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$ (From (i) and (ii)

So, $a^2+b^2=81+45=126$

$[x{]}^2+2[x+2]-7=0$

$\begin{array}{rr}& \Rightarrow [x{]}^2+2[x]+4-7=0\\ & \Rightarrow \left[x\right]=1,-3\\ & \Rightarrow x\in \left[\mathrm{1,2}\right)\cup [-3,-2)\end{array}$

27. Since the parabola is symmetric with respect to y-axis and has vertex (0, 0)

The equation in of the form x³ = 4ay or x² = 4ay.

The parabola passes through (5, 2) which lies on the 1st quadrant

? The equation of parabola is of the form,

x² = 4ay

Putting x = 5 and y = 2,

(5)² = 4 (a) (2)

25 = 8a

a = $\frac{25}{8}$

? The equation of the parabola is,

x² = 4ay

$\Rightarrow x^2=4\left(\frac{25}{8}\right)y$

$\Rightarrow x^2=\frac{25}{2}y$

2x² = 25y

26. Since the axis of parabola is x-axis,

The equation parabola is either y² = 4ax or y² = 4ax.

Also it passes through (2, 3) which lies in the first quadrant.

So the equation is,

y²= 4ax

Putting x = 2 and y = 3, we set

(3)² = 4 (a) (2)

a = $\frac{9}{8}$

? The equation of parabola is y² = 4 $*\frac{9}{8}x$

y² = $\frac{9}{2}x$

2y² = 9x

25. Focus (-2, 0) lies on x-axis and the x-Co-ordinate is negative.

The equation must be, y² = 4ax

Co-ordinate of focus = (–a, 0) = (–2, 0)

–a = –2

a = 2

? Equation of a parabola is,

y² = –4 (2)x

y² = –8x