The axis of a parabola is the line $ y = x $ and its vertex and focus are in the first quadrant at distances $ \sqrt{2} $ and $ 2\sqrt{2} $ units from the origin, respectively. If the point $ (1, k) $ lies on the parabola, then a possible value of $ k $ is:
Show Hint
- 4
- 9
- 3
- 8
The Correct Option is B
Approach Solution - 1
The vertex of the parabola is at the origin \( (0, 0) \), and the axis of the parabola is along the line \( y = x \). The focus is at \( (2\sqrt{2}, 2\sqrt{2}) \), and the directrix is the line \( x + y = 0 \).
Using the definition of a parabola, the distance from any point on the parabola to the focus equals the distance from that point to the directrix. Let the point \( P(1, k) \) be on the parabola.
Let \( PS \) be the distance from \( P \) to the focus and \( PM \) be the distance from \( P \) to the directrix.
First, calculate the distance \( PS \): \[ PS = \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} \] Next, calculate the distance \( PM \) from the point \( P(1, k) \) to the directrix \( x + y = 0 \).
The formula for the distance from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is: \[ PM = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \] For the directrix \( x + y = 0 \), \( a = 1 \), \( b = 1 \), and \( c = 0 \), so: \[ PM = \frac{|1 \times 1 + k|}{\sqrt{1^2 + 1^2}} = \frac{|1 + k|}{\sqrt{2}} \] Now, equate \( PS \) and \( PM \) (since the point lies on the parabola): \[ \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} = \frac{|1 + k|}{\sqrt{2}} \] After solving this equation, we find that \( k = 9 \).
Thus, the correct answer is \( 9 \).
Approach Solution -2
Given: The equation of the directrix is \( x + y = 0 \), and the condition is \( PS = PM \), where \( P \) is a point on the curve, and \( S \) and \( M \) are points related to the directrix and the focus.
Step 1: Express the distance from \( P \) to the directrix and from \( P \) to the focus.
The distance from \( P \) to the directrix is:
\[ \sqrt{(1 - 2)^2 + (K - 2)^2} = \frac{(1 + K)}{\sqrt{2}}. \]Step 2: Simplify and solve the equation:
\[ 2K^2 + 8K - 8K + 2 = K^2 + 1 + 2K. \]This simplifies to:
\[ K^2 - 10K + 9 = 0. \]Step 3: Solve the quadratic equation \( K^2 - 10K + 9 = 0 \) to get:
\[ K = 9. \]Final Answer: \( K = 9 \).
Top Questions on Geometry
- Ronny uses a 5-digit key for a combination lock, where 5 digits need to be entered in a fixed sequence. While he remembers that the 5 digits are 9, 8, 7, 5, and 4, he has forgotten the sequence he uses. He also remembers that the sum of the first three digits is a multiple of 3, and so is the sum of the last three digits. Further, the sum of the last four digits is a multiple of 4.
Which of the following is DEFINITELY FALSE?
- During Durga Puja, for the purpose of lighting, one puja pandal in Kolkata used many identical structures made of wooden sticks. The design of the structures was as follows: each structure was constructed with the help of six wooden sticks by combining an isosceles triangular structure, and a square structure, with the bases of both structures being the same. Let us take one such structure. Call the triangle PAB, with PA = PB, and the square ABCD, with AB being the same wooden stick as a common base for the triangle and the square. To make the structure strong, the two equal sides of the triangular structure were tied with the opposite side of square’s base, i.e., CD, at points E and F, in such a way that CE = EF = FD. The structure was hung from P.
If AB = 0.5m (meter), the total length of wooden sticks required for twenty such structures is:
- Rajan is a fruit seller. On any day, he sells only one kind of fruit. On the first day, he buys 9 kg of blueberries. On the second day, he buys 22 kg of kiwis. On the third day, he buys mangoes at Rs. 35/kg and spends Rs. 15 less than any of the previous three days.
If he then sells all the mangoes at Rs. 50/kg, what is his MINIMUM possible profit on the fourth day?
- During Durga Pooja, six wooden sticks are used to create a structure to be used in a Pandal. First, an isosceles triangle is created using three sticks. Let this triangle be \(PAB\) with \(PA = PB\). Then a square \(ABCD\) is completed using the common base \(AB\). To strengthen this structure, the equal sides \(PA\) and \(PB\) are also joined to points \(E\) and \(F\) on side \(CD\) such that \(CE = EF = FD\). What is the total length of sticks used to create 20 such structures if \(AB = 0.5\) meters?
- Two equal circles of radius 5 cm are touching each other at point \(P\). A common tangent to the circles touches them at points \(Q\) and \(R\). A square is drawn inside triangle \(PQR\) such that one of its sides is resting on this common tangent and the other two vertices touch the two circles. Find the area of the square.
Questions Asked in JEE Main exam
- The number of non-empty equivalence relations on the set \(\{1,2,3\}\) is :
- JEE Main - 2025
- Set Theory
- A liquid when kept inside a thermally insulated closed vessel at \(25^{\circ}C\) was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?
- JEE Main - 2025
- Thermodynamics
- The value of \[ \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\left( (\log_e x)^2 +1 \right)^{-1}}}{e^{\left( (\log_e x)^2 +1 \right)^{-1}} + e^{\left( (6-\log_e x)^2 +1 \right)^{-1}}} \right) dx \] is:
- JEE Main - 2025
- Integration
In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to \( R_1 \) and \( R_2 \), i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is:
- A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let \( r \) be the radius of a circle that has centre at the point \( (2, 5) \) and intersects the circle C at exactly two points. If the set of all possible values of \( r \) is the interval \( (\alpha, \beta) \), then \( 3\beta - 2\alpha \) is equal to:
- JEE Main - 2025
- Coordinate Geometry