Conic Sections

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

24. Since the focus (3, 0) lies on the x-axis, the x-axis is the axis of parabola.

Hence the equation is either y² = 4ax or y² = –4ax

Since the focus has positive x co-ordinate,

The equation must be y² = 4ax

Co-ordinate of focus (a, 0) = (3, 0)

a = 3

? The equation is given by

y² = 4 (3)x

y² = 12x

23. Since the focus (0, –3) lies on the y–axis, the y–axis is the axis of parabola.

Hence the equation is either x² = 4ay or x² = –4ay

Since the directrix is y = 3 and the forces (0, –3) has negative y Co–ordinate.

The equation must be

x² = –4ay

Co–ordinate of focus = (0, –a)

(0, –a) = (0, –3)

–a = –3

a = 3

? Equation of parabola is

x² = –4ay

x² = –4 (3)y

x² = –12y

22. Since the focus (6, 0) lien on the x–axis, the x–axis is the axis of parabola

Hence the equation is either y² = 4ax or y² = –4ax

Since the directrix is x = –6 and the focus (6, 0) has positive x – Coordinate,

The equation must be y² = 4ax

with a = 6

Hence, equation of parabola is,

y² = 4ax

y² = 4 (6)x

y² = 24x

18.  y² = –4ax

Comparing with the given equation y² = –8x

We get,

–4ax = –8x

a = $\frac{8}{4}=2$

? Co–ordinates of focus is (–0, 0) = (–2, 0)

Axis of Parabola : x-axis.

Equation of directrix is,

x = a

x = 2

Length of latus rectum =4a

= 4 * 2 = 8