Find the coordinates of the vertex, focus, and the equations of the directrix and axes of the parabolasi) y2−x+4y+5=0 ii) 3x2−9x+5y−2=0 iii) x2+8x+12y+4=0

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Find the coordinates of the vertex, focus, and the equations of the directrix and axes of the parabolas
$i) y^{2} - x + 4 y + 5 = 0 ii) 3 x^{2} - 9 x + 5 y - 2 = 0 iii) x^{2} + 8 x + 12 y + 4 = 0$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$i) Given parabola y^{2} - x + 4 y + 5 = 0$
$\begin{array}{l} y^{2} + 4 y + 4 - 4 = x - 5 \\ (y + 2)^{2} = x - 5 + 4 \Rightarrow (y + 2)^{2} = 1 (x - 1) \end{array}$
Compare $(y - k)^{2} = 4 a (x - h)$
$4 a = 1 \Rightarrow a = \frac{1}{4}$
$Vertex A = (h, k) = (1, - 2)$
$Focus S = (h + a, k) = (1 + \frac{1}{4}, - 2) = (\frac{5}{4}, - 2)$
Equation of directrix of x - h + a = 0
$x - 1 + \frac{1}{4} = 0 \Rightarrow 4 x - 4 + 1 = 0 \Rightarrow 4 x - 3 = 0$
$Equation of axis is y - k = 0 \Rightarrow y + 2 = 0$
$ii) Given parabola is 3 x^{2} - 9 x + 5 y - 2 = 0$
$\begin{array}{l} 3 (x^{2} - 3 x) = - 5 y + 2 \\ 3 {(x - \frac{3}{2})}^{2} = - 5 y + 2 + \frac{27}{4} = - 5 (y - \frac{7}{4}) \\ {(x - \frac{3}{2})}^{2} = - \frac{5}{3} (y - \frac{7}{4}) \end{array}$
$Compare with (x - h)^{2} = - 4 a (y - k)$
$4 a = \frac{5}{3} \Rightarrow a = \frac{5}{12}$
$Vertex A = (h, k) = (\frac{3}{2}, \frac{7}{4}) Focus S = (h, k - a) = (\frac{3}{2}, \frac{7}{4} - \frac{5}{12}) = (\frac{3}{2}, \frac{4}{3}) Equation of directrix is y - k - a = 0 y - \frac{7}{4} - \frac{5}{12} = 0 \Rightarrow 6 y = 13 Equation of axis is x - h = 0 x - \frac{3}{2} = 0 \Rightarrow 2 x - 3 = 0$
$iii) Given parabola is x^{2} + 8 x + 12 y + 4 = 0 \begin{array}{l} x^{2} + 8 x + 16 - 16 + 12 y + 4 = 0 \\ (x + 4)^{2} = - 12 y + 12 = - 12 (y - 1) Compare with (x - h)^{2} = - 4 a (y - k) 4 a = 12 \Rightarrow a = 3 \end{array}$
$Vertex A = (h, k) = (- 4, 1) Focus S = (h, k - a) = (- 4, 1 - 3) = (- 4, - 2) Equation of directrix is y - k - a = 0 y - 1 - 3 = 0 \Rightarrow y - 4 = 0 Equation of axis is x - h = 0 \Rightarrow x + 4 = 0$