Question:
The focus of the parabola $y^2 - 4y - x+3=0$ is
The focus of the parabola $y^2 - 4y - x+3=0$ is
Updated On: Jun 7, 2024
- $\left( \frac{3}{4} , 2 \right)$
- $\left( \frac{3}{4} , - 2 \right)$
- $\left( 2 , \frac{3}{4} \right)$
- $\left( \frac{-3}{4} , 2 \right)$
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
$y^{2}-4 y-x+3=0$
$(y-2)^{2}-4-x+3 =0$
$(y-2)^{2}-x-1 =0$
$(y-2)^{2} =(x+1)$
Let $Y^{2}=X$ ...(i)
Here, $Y=(y-2), X=(x+1)$
Vertices $(X=0, Y=0)=(2,-1)$
E (i) comparing on $y^{2}=4 a x$
$4 a=1$
$\Rightarrow \, a=\frac{1}{4}$
$\therefore$ Focus $=\left(\frac{1}{4}-1,2\right)=\left(-\frac{3}{4}, 2\right)$
Was this answer helpful?
1
2
Top Questions on Parabola
- Let \( L_1, L_2 \) be the lines passing through the point \( P(0, 1) \) and touching the parabola \[ 9x^2 + 12x + 18y - 14 = 0. \] Let \( Q \) and \( R \) be the points on the lines \( L_1 \) and \( L_2 \) such that the \( \Delta PQR \) is an isosceles triangle with base \( QR \). If the slopes of the lines \( QR \) are \( m_1 \) and \( m_2 \), then \( 16\left(m_1^2 + m_2^2\right) \) is equal to ______.
- Let \( C \) be the circle of minimum area touching the parabola \( y = 6 - x^2 \) and the lines \( y = \sqrt{3} |x| \). Then, which one of the following points lies on the circle \( C \)?
- Let the length of the focal chord \( PQ \) of the parabola \( y^2 = 12x \) be 15 units. If the distance of \( PQ \) from the origin is \( p \), then \( 10p^2 \) is equal to _____
- Let \( PQ \) be a chord of the parabola \( y^2 = 12x \) and the midpoint of \( PQ \) be at \( (4, 1) \). Then, which of the following points lies on the line passing through the points \( P \) and \( Q \)?
- The area (in square units) of the region described by: \[ \{(x, y) : y^2 \leq 2x, \, \text{and} \, y \geq 4x - 1\} \] is:
View More Questions
Questions Asked in KEAM exam
- A Spherical ball is subjected to a pressure of 100 atmosphere. If the bulk modulus of the ball is 1011 N/m2,then change in the volume is:
- KEAM - 2023
- mechanical properties of solids
- Two identical solid spheres,each of radius 10cm,are kept in contact. If the mass of inertia of this system about the tangent passing through the point of contact is 0. Then mass of each sphere is:
- KEAM - 2023
- System of Particles & Rotational Motion
- The value of a(≠0) for which the equation \(\frac{1}{2}(x-2)^2+1=\sin(\frac{a}{x})\) holds is/are
- KEAM - 2023
- Trigonometric Functions
- \(∫xlog(1+x^2)dx=\)?
- KEAM - 2023
- Integration by Parts
- A uniform thin rod of mass 3kg has a length of 1m. If a point mass of 1kg is attached to it at a distance of 40cm from its center,the center of mass shifts by a distance of:
- KEAM - 2023
- System of Particles & Rotational Motion
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.