Question:
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
Updated On: Apr 8, 2024
- y = 4(x – 2)
- y = 4(x – 1)
- y = 4(x + 1)
- y = 4(x + 2)
Hide Solution
Verified By Collegedunia
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
\(y = 4x-4\)
Was this answer helpful?
0
0
Top Questions on Parabola
- 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:
- If the line \( 3x - 2y + 12 = 0 \) intersects the parabola \( 4y = 3x^2 \) at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle 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:
- 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:
- 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 points A and B, then \( (AB)^2 \) is equal to:
View More Questions
Questions Asked in JEE Main exam
- Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of \( \text{ClF}_3 \). The ions from the following with ‘n’ number of unpaired electrons are:
A. \( \text{V}^{3+} \)
B. \( \text{Ti}^{3+} \)
C. \( \text{Cu}^{2+} \)
D. \( \text{Ni}^{2+} \)
E. \( \text{Ti}^{2+} \)
Choose the correct answer from the options given below:- JEE Main - 2025
- Thermodynamics
- The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet color in non-luminous flame under hot condition in borax bead test is:
- JEE Main - 2025
- coordination compounds
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Find the equivalent capacitance between A and B, where \( C = 16 \, \mu F \).
- JEE Main - 2025
- Capacitors
If the equation of the parabola with vertex \( \left( \frac{3}{2}, 3 \right) \) and the directrix \( x + 2y = 0 \) is \[ ax^2 + b y^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} \]
- JEE Main - 2025
- Calculus
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.