Conic Sections

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol: 

Sol:

$Givenequationare3x-4y+4=0\\ and6x-8y-7=0\Rightarrow 3x-4y-\frac{7}{2}=0$
$\begin{array}{l}\text{Since}\frac{3}{6}=\frac{-4}{-8}=\frac{1}{2}thenthelinesareparallel.\\ So,thebetweentheparallellines=\left|\frac{c_1-c_2}{\sqrt[]{a^2+b^2}}\right|=\left|\frac{4+\frac{7}{2}}{\sqrt[]{{\left(3\right)}^2+{\left(-4\right)}^2}}\right|\\ =\left|\frac{\frac{15}{2}}{5}\right|=\frac{3}{2}\\ Diameter=\frac{3}{2}\\ \therefore Radius=\frac{3}{4}\\ Hence,therequiredradius=\frac{3}{4}.\end{array}$

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 =&

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 $\times \frac{9}{8}x$

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 =

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 (

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

18. Given,

h = $\frac{1}{2}$ , k = $\frac{1}{4}$ , r = $\frac{1}{12}$

? Equation of the circle is,

(x – h)² + (y – k)² = r²

${\left(x-\frac{1}{2}\right)}^2+{\left\{y-\left(\frac{1}{4}\right)\right\}}^2={\left(\frac{1}{12}\right)}^2$

${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{1}{4}\right)}^2=\frac{1}{144}$

$\Rightarrow 144x{}_2+144y^2-144x-72y+44=0$

17. Centre (–2, 3) and radius 4.

Given, h = –2, k = 3, r = 4

? Equation of the circle is,

(x – h)² + (y – k)² = 2²

(x + 2)² + (y – 3)² = 16

$\Rightarrow x^2+y^2+4x-6y-3=0$