Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = _______.
Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = _______.
Correct Answer: 10
Solution and Explanation
The correct answer is 10
Focus = (4, 4) and vertex = (3, 2)
∴ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line
x + 2y = 6 is B(x2, y2)
\(∴\frac{x^2−2}{1}=\frac{y^2−0}{2}=\frac{−2(2+0−6)}{5}\)
\(∴B(x_2,y_2)=(185,165).\)
Point B is point of intersection of direction with axes of parabola P2.
∴ x + 2y = λ must have point
\((\frac{18}{5},\frac{16}{5})\)
∴ x + 2y = 10
Top Questions on Parabola
- An equilateral triangle OAB is inscribed in the parabola $y^2 = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is
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:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
- 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 the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Questions Asked in JEE Main exam
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
- JEE Main - 2026
- Coordinate Geometry
- A voltage regulating circuit consisting of a Zener diode having breakdown voltage of $10\,\text{V}$ and maximum power dissipation of $0.4\,\text{W}$ is operated at $15\,\text{V}$. The approximate value of protective resistance in this circuit is ___ $\Omega$.
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
- Two point charges of \(1\,\text{nC}\) and \(2\,\text{nC}\) are placed at two corners of an equilateral triangle of side \(3\) cm. The work done in bringing a charge of \(3\,\text{nC}\) from infinity to the third corner of the triangle is ________ \(\mu\text{J}\). \[ \left(\frac{1}{4\pi\varepsilon_0}=9\times10^9\,\text{N m}^2\text{C}^{-2}\right) \]
- JEE Main - 2026
- Electrostatics
Consider the following two reactions A and B:
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
- JEE Main - 2026
- Organic Chemistry
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
- JEE Main - 2026
- Arithmetic Progression
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.