Find the equation of each of the following parabolas:

Directrix $x = 0$ , focus at (6, 0).

Vertex at (0, 4), focus at (0, 2).

Focus at (-1,-2), directrix $x - 2 y + 3 = 0$ .

2 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

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

Sol:

$\begin{array}{l} (i) G i v e n t h a t d i r e c t r i x = 0 a n d f o c u s (6, 0) \\ ∴ T h e e q u a t i o n o f t h e p a r a b o l a i s \\ {(x - 6)}^{2} + y^{2} = x^{2} \\ \Rightarrow x^{2} + 36 - 12 x + y^{2} = x^{2} \\ \Rightarrow y^{2} - 12 x + 36 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s y^{2} - 12 x + 36 = 0 \\ (i i) G i v e n t h a t v e r t e x a t (0, 4) a n d f o c u s a t (0, 2) . \\ S o, e q u a t i o n o f d i r e c t r i x i s y - 6 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | y - 6 | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 4 - 4 y} = | y - 6 | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} + 4 - 4 y = 36 - 12 y \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ \Rightarrow x^{2} = 32 - 8 y \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} = 32 - 8 y \\ (i i i) G i v e n t h a t f o c u s a t (- 1, - 2) a n d d i r e c t r i x x - 2 y + 3 = 0 \\ L e t (x, y) b e a n y o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x + 1)}^{2} + {(y + 2)}^{2}} = | \frac{x - 2 y + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + 1 + 2 x + y^{2} + 4 + 4 y} = | \frac{x - 2 y + 3}{\sqrt[]{5}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} + 4 + 4 y = \frac{x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x}{5} \\ \Rightarrow 5 x^{2} + 5 + 10 x + 5 y^{2} + 20 + 20 y = x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x \\ \Rightarrow 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 . \end{array}$

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

Alok Kumar Singh

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Alok Kumar Singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Alok Kumar Singh

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

Alok Kumar Singh

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

Alok Kumar Singh

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$