A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with
- Length of latus rectum 3
- Length of latus rectum 6
- \(Focus \bigg(\frac{4}{3},0\bigg)\)
- \(Focus \bigg(0,\frac{3}{4}\bigg)\)
The Correct Option is A
Solution and Explanation
\(2x-y \frac{dx}{dy} = 0\)
tangent at \(P:y-y =\frac{dy}{dx}(y-x)\)
\(∴ 2 \frac{dy}{y} = \frac{dx}{x}\)
\(⇒ 2Iny = Inx+Inc\)
\(⇒ y^2 = cx\)
At coordinates \((3, 3)\) curves pass through
Hence, \(c = 3\)
Therefore, the is parabola :
\(y^2 = 3x \)
So Length of latus rectum is \(3\).
Hence, the correct option is (A): Length of latus rectum is \(3\)
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:
Questions Asked in JEE Main exam
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
- JEE Main - 2025
- Conic sections
- Let \( \alpha_1 \) and \( \beta_1 \) be the distinct roots of \( 2x^2 + (\cos\theta)x - 1 = 0, \ \theta \in (0, 2\pi) \). If \( m \) and \( M \) are the minimum and the maximum values of \( \alpha_1 + \beta_1 \), then \( 16(M + m) \) equals:
- JEE Main - 2025
- Maxima and Minima
- Let the line \( x + y = 1 \) meet the circle \( x^2 + y^2 = 4 \) at the points A and B. If the line perpendicular to AB and passing through the midpoint of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to:
- JEE Main - 2025
- Coordinate Geometry
- The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is:
- JEE Main - 2025
- permutations and combinations
- Two identical symmetric double convex lenses of focal length \( f \) are cut into two equal parts \( L_1, L_2 \) by the AB plane and \( L_3, L_4 \) by the XY plane as shown in the figure respectively. The ratio of focal lengths of lenses \( L_1 \) and \( L_3 \) is:
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.