Home
Class 12
MATHS
If for a conic section a focus is (-1,1)...

If for a conic section a focus is (-1,1), eccentricity=3 and the equation of the corresponding directrix is x-y+3=0, then the equation of this conic section is

A

`7x^(2)-18xy+7y^(2)+50x-50y+77=0`

B

`7x^(2)+18xy+7y^(2)=1`

C

`7x^(2)+18xy+7y^(2)-50x+50y+77=0`

D

none

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation of the conic section given the focus, eccentricity, and directrix, we can follow these steps: ### Step 1: Identify the given values - Focus (F) = (-1, 1) - Eccentricity (e) = 3 - Directrix: \( x - y + 3 = 0 \) ### Step 2: Set up the point P Let P be a point on the conic section with coordinates (x, y). ### Step 3: Calculate the distance from P to the focus F The distance \( PF \) from point P to the focus F is given by the distance formula: \[ PF = \sqrt{(x + 1)^2 + (y - 1)^2} \] ### Step 4: Calculate the distance from P to the directrix The distance \( PD \) from point P to the directrix can be calculated using the formula for the distance from a point to a line \( Ax + By + C = 0 \): \[ PD = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} \] For the directrix \( x - y + 3 = 0 \), we have: - \( A = 1 \), \( B = -1 \), \( C = 3 \) Thus, \[ PD = \frac{|1 \cdot x - 1 \cdot y + 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 3|}{\sqrt{2}} \] ### Step 5: Set up the equation based on the definition of conics According to the definition of a conic section, the distance from the point P to the focus is equal to the eccentricity times the distance from P to the directrix: \[ PF = e \cdot PD \] Substituting the expressions we found: \[ \sqrt{(x + 1)^2 + (y - 1)^2} = 3 \cdot \frac{|x - y + 3|}{\sqrt{2}} \] ### Step 6: Square both sides to eliminate the square root Squaring both sides gives: \[ (x + 1)^2 + (y - 1)^2 = 9 \cdot \frac{(x - y + 3)^2}{2} \] ### Step 7: Expand both sides Expanding the left side: \[ (x + 1)^2 + (y - 1)^2 = x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 + y^2 + 2x - 2y + 2 \] Expanding the right side: \[ 9 \cdot \frac{(x - y + 3)^2}{2} = \frac{9}{2} (x^2 - 2xy + y^2 + 6x - 6y + 9) = \frac{9}{2} x^2 - 9xy + \frac{9}{2} y^2 + 27x - 27y + \frac{81}{2} \] ### Step 8: Combine and simplify Setting both sides equal: \[ x^2 + y^2 + 2x - 2y + 2 = \frac{9}{2} x^2 - 9xy + \frac{9}{2} y^2 + 27x - 27y + \frac{81}{2} \] Bringing all terms to one side gives: \[ 0 = \frac{9}{2} x^2 - x^2 - 9xy + \frac{9}{2} y^2 - y^2 + 27x - 2x - 27y + 2 + \frac{81}{2} \] This simplifies to: \[ \frac{7}{2} x^2 + \frac{7}{2} y^2 - 9xy + 25x - 25y + \frac{85}{2} = 0 \] ### Step 9: Multiply through by 2 to eliminate fractions Multiplying through by 2 gives: \[ 7x^2 + 7y^2 - 18xy + 50x - 50y + 85 = 0 \] ### Final Equation Thus, the equation of the conic section is: \[ 7x^2 + 7y^2 - 18xy + 50x - 50y + 85 = 0 \]
Promotional Banner

Topper's Solved these Questions

  • CONIC SECTIONS

    MTG-WBJEE|Exercise WB JEE WORKOUT (CATEGORY 3 : One or More than One Option Correct Type)|15 Videos
  • CONIC SECTIONS

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (CATEGORY 1 : Single Option Correct Type)|32 Videos
  • CONIC SECTIONS

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (CATEGORY 3 : One or More than One Option Correct Type)|6 Videos
  • COMPLEX NUMBERS

    MTG-WBJEE|Exercise WB JEE Previous Years Questions (CATEGORY 3 : One or More than One Option Correct Type (2 Marks) )|3 Videos
  • DEFINITE INTEGRALS

    MTG-WBJEE|Exercise WE JEE PREVIOUS YEARS QUESTIONS (CATEGORY 3 : ONE OR MORE THAN ONE OPTION CORRECT TYPE)|5 Videos

Similar Questions

Explore conceptually related problems

If 7 x^2 + p xy + q y^2 + rx - sy + t = 0 is the equation of the hyperbola whose one focus is (-1,1), eccentricity = 3 and the equation of corresponding directrix is x-y+3 = 0, then value of 't' is :

Conic Section L3

Conic Section L1

Find the equation of the hyperbola whose one focus is (-1, 1) , eccentricity = 3 and the equation of the corresponding directrix is x - y + 3 = 0 .

If the focus =(2,3)and directrix is x+y=1 then the equation of the parabola is ____.

If for a rectangular hyperbola a focus is (1,2) and the corresponding directrix is x+y=1 then the equation of the rectangular hyperbola is

Find the equation of the hyperbola whose focus is (1,1), eccentricity is 2 and equation of directrix is x+y+1=0.

Find the equation of the parabola with focus (-3,0) and the equation of the directrix is x = 3.

Conic Section: Lecture 3

Conic Section: Lecture 3