Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of π/4 with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is
- 8 only
- 2 only
\(\frac{1}{4}\) only
- any a > 0
The Correct Option is D
Solution and Explanation
The correct answer is (D) : any a > 0
\(P:y²=4ax, a > 0 \) \(S(a,0)\)
Equation of tangent on parabola
\(y = mx + \frac{a}{m}\)
y = 3x + 5
\(tan \frac{π}{4} = |\frac{m-3}{1+3m} | ⇒ m-3 = ± ( 1+3m )\)
m-3 = 1+3m
m=-2
m-3 = -1-3m
\(m=\frac{1}{2}\)
Equation of one tangent :\( y = -2x - \frac{a}{2}\)
Equation of other tangent : \(y = \frac{x}{2} + 2a\)
Point of contact are
\(( \frac{a}{(-2)²} , \frac{-2a}{(-2)} ) and ( \frac{a}{(\frac{1}{2})²} , \frac{-2a}{\frac{1}{2}} )\)
\(A ( \frac{a}{4},a)\) and \(B (4a,-4a)\)
Now or \((ΔABS)\) = 0 [ S is the focus ]
\(\frac{1}{2} \begin{vmatrix} \frac{a}{4} & a & 1 \\ 4a & -4a & 1 \\ a& 0 & 1 \end{vmatrix} = 0\)
\(⇒ \frac{a}{4} (-4a-0) -a(4a-a) + 1(0-(-4a²)) = 0\)
\(= -a² -3a² + 4a² = 0\)
Always true
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
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\).
- JEE Main - 2026
- Current electricity
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
- Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are \(2.4 \times 10^8\) m/s and \(2.7 \times 10^8\) m/s, respectively, then the value of critical angle is :
- JEE Main - 2026
- Newtons Laws of Motion
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is:
- JEE Main - 2026
- Thermodynamics
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
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.