The equation of a common tangent to the parabolas y = x2 and y = –(x – 2)2 is
The equation of a common tangent to the parabolas y = x2 and y = –(x – 2)2 is
- y = 4(x – 2)
- y = 4(x – 1)
- y = 4(x + 1)
- y = 4(x + 2)
The Correct Option is B
Solution and Explanation
The correct answer is (B) : y = 4(x – 1)
Equation of tangent of slope m to y = x2
\(y=mx−\frac{1}{4}m^2\)
Equation of tangent of slope m to y = –(x – 2)2
\(y=m(x−2)+\frac{1}{4}m^2\)
If both equation represent the same line
\(\frac{1}{4}m^2−2m=−\frac{1}{4}m^2\)
m = 0, 4
So, equation of tangent
\(y = 4x-4\)
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
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 \( y^2 = 12x \) 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 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
- Two tuning forks \(A\) and \(B\) are sounded together giving rise to 8 beats in 2 s. When fork \(A\) is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork \(B\) is \(380\ \text{Hz}\), find the original frequency of tuning fork \(A\). Given: \[ f_B = 380\ \text{Hz} \]
- JEE Main - 2026
- General Physics
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
- JEE Main - 2026
- Three Dimensional Geometry
- Let \[ A=\{z\in\mathbb{C}:|z-2|\le 4\} \quad\text{and}\quad B=\{z\in\mathbb{C}:|z-2|+|z+2|=5\}. \] Then the maximum value of \(|z_1-z_2|\), where \(z_1\in A\) and \(z_2\in B\), is:
- JEE Main - 2026
- Miscellaneous
- Let \(y=y(x)\) be the solution of the differential equation \[ x\frac{dy}{dx}=y-x^2\cot x,\quad x\in(0,\pi) \] If \(y\!\left(\frac{\pi}{2}\right)=\frac{\pi^2}{2}\), then \[ 6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) \] is equal to:
- JEE Main - 2026
- Miscellaneous
- The sum of the coefficients of \(x^{499}\) and \(x^{500}\) in \[ (1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000} \] is:
- JEE Main - 2026
- Binomial theorem
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.