Question:
For the parabola $y^2 + 6\,y - 2x = - 5$ I. the vertex is $( - 2,-3)$ II. the directrix is $y + 3 = 0$ Which of the following is correct?
For the parabola $y^2 + 6\,y - 2x = - 5$ I. the vertex is $( - 2,-3)$ II. the directrix is $y + 3 = 0$ Which of the following is correct?
Updated On: Mar 11, 2025
- Both I and II are correct
- I is true, II is false
- Both I and II are false
- I is false, II is true
Hide Solution
Shivam
The Correct Option is B
Solution and Explanation
We have,
$y^{2}+6 \,y-2\, x=-5$
$\Rightarrow y^{2}+6\, y=2\, x-5$
$\Rightarrow y^{2}+6\, y+9=2 x-5+9$
$\Rightarrow (y+3)^{2}=2\, x+4$
$\Rightarrow (y+3)^{2}=2(x+2)$
$\therefore 4 a=2 \Rightarrow a=\frac{1}{2}$
Vertex $=(-2,-3)$
Equations of directrix
$x+2=-\frac{1}{2}$
$\Rightarrow x+\frac{5}{2}=0$
$\Rightarrow 2 \,x+5=0$
$y^{2}+6 \,y-2\, x=-5$
$\Rightarrow y^{2}+6\, y=2\, x-5$
$\Rightarrow y^{2}+6\, y+9=2 x-5+9$
$\Rightarrow (y+3)^{2}=2\, x+4$
$\Rightarrow (y+3)^{2}=2(x+2)$
$\therefore 4 a=2 \Rightarrow a=\frac{1}{2}$
Vertex $=(-2,-3)$
Equations of directrix
$x+2=-\frac{1}{2}$
$\Rightarrow x+\frac{5}{2}=0$
$\Rightarrow 2 \,x+5=0$
Was this answer helpful?
0
1
Top Questions on Parabola
- Let the point $ P $ of the focal chord $ PQ $ of the parabola $ y^2 = 16x $ be $ (1, -4) $. If the focus of the parabola divides the chord $ PQ $ in the ratio $ m : n $, gcd($m, n$) = 1, then $ m^2 + n^2 $ is equal to:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
- Let \( P(4, 4\sqrt{3}) \) be a point on the parabola \( y^2 = 4ax \) and PQ be a focal chord of the parabola. If M and N are the foot of the perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
- Let A and B be the two points of intersection of the line \( y + 5 = 0 \) and the mirror image of the parabola \( y^2 = 4x \) with respect to the line \( x + y + 4 = 0 \). If \( d \) denotes the distance between A and B, and \( a \) denotes the area of \( \Delta SAB \), where \( S \) is the focus of the parabola \( y^2 = 4x \), then the value of \( (a + d) \) is:
View More Questions
Questions Asked in TS EAMCET exam
- $7\hat{i}-4\hat{j}+7\hat{k}, \hat{i}-6\hat{j}+10\hat{k}, -\hat{i}-3\hat{j}+4\hat{k}, 5\hat{i}-\hat{j}+\hat{k}$ are the position vectors of the points A, B, C, D respectively. If $p\hat{i} + q\hat{j} + r\hat{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral ABCD, then $p+q+r=$
- TS EAMCET - 2025
- Vector Algebra
- A particle is executing simple harmonic motion with an amplitude of 10 cm. If the kinetic energy of the particle at a distance of 6 cm from the mean position is 100 J, then the kinetic energy of the particle at a distance of 2 cm from the mean position is
- TS EAMCET - 2025
- Waves and Oscillations
- If the sum of two vectors is a unit vector, then the magnitude of their difference is:
- Which of the following is an example of antifertility drug?
- TS EAMCET - 2025
- Chemistry in Everyday Life
In a messenger RNA molecule, untranslated regions (UTRs) are present at:
I. 5' end before start codon
II. 3' end after stop codon
III. 3' end before stop codon
IV. 5' end after start codon- TS EAMCET - 2025
- Plant Ecology
View More Questions
Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.