Home
Class 12
MATHS
If e(1) and e(2) are the eccentricites...

If `e_(1)` and `e_(2)` are the eccentricites of `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and `(x^(2))/(b^(2))+(y^(2))/(a^(2))=1` respectively then

A

`e_(1)=e_(2)`

B

`e_(1)e_(2)=1`

C

`(1)/(e_(1)^(2))+(1)/(e_(2)^(2))=1`

D

`(1)/(e_(1))+(1)/(e_(2))=1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the eccentricities \( e_1 \) and \( e_2 \) of the two given ellipses and then compare them. ### Step 1: Identify the equations of the ellipses The equations given are: 1. \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (Equation 1) 2. \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) (Equation 2) ### Step 2: Find the eccentricity \( e_1 \) of the first ellipse For the first ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): - The formula for the eccentricity \( e \) of an ellipse is given by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] - Here, \( e_1 = \sqrt{1 - \frac{b^2}{a^2}} \). ### Step 3: Find the eccentricity \( e_2 \) of the second ellipse For the second ellipse \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \): - Using the same formula for eccentricity: \[ e_2 = \sqrt{1 - \frac{a^2}{b^2}} \] ### Step 4: Equate the two eccentricities Now we have: 1. \( e_1 = \sqrt{1 - \frac{b^2}{a^2}} \) 2. \( e_2 = \sqrt{1 - \frac{a^2}{b^2}} \) To compare \( e_1 \) and \( e_2 \), we can square both sides: \[ e_1^2 = 1 - \frac{b^2}{a^2} \] \[ e_2^2 = 1 - \frac{a^2}{b^2} \] ### Step 5: Set the equations equal to each other Setting \( e_1^2 \) equal to \( e_2^2 \): \[ 1 - \frac{b^2}{a^2} = 1 - \frac{a^2}{b^2} \] ### Step 6: Simplify the equation Canceling out the 1's gives: \[ -\frac{b^2}{a^2} = -\frac{a^2}{b^2} \] Multiplying through by \( -1 \): \[ \frac{b^2}{a^2} = \frac{a^2}{b^2} \] ### Step 7: Cross-multiply Cross-multiplying gives: \[ b^4 = a^4 \] ### Step 8: Take the square root Taking the square root of both sides gives: \[ b^2 = a^2 \] This implies \( b = a \) or \( b = -a \). However, since \( a \) and \( b \) are lengths, we take \( b = a \). ### Conclusion Thus, we find that: \[ e_1 = e_2 \] Therefore, the eccentricities are equal. ### Final Answer The correct option is that \( e_1 = e_2 \). ---
Promotional Banner

Similar Questions

Explore conceptually related problems

If e_(1)ande_(2) be the eccentricities of the ellipses (x^(2))/(a^(2))+(y^(2))/(b^(2))=1and(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1 respectively then prove that 3=4e_(2)^(2)-e_(1)^(2) .

Statement- 1 : If 5//3 is the eccentricity of a hyperbola, then the eccentricity of its conjugate hyperbola is 5//4 . Statement- 2 : If e and e' are the eccentricities of hyperbolas (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and (x^(2))/(a^(2))-(y^(2))/(b^(2))=-1 respectively, then (1)/(e^(2))+(1)/(e'^(2))=1 .

For the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 and (x^(2))/(b^(2))+(y^(2))/(a^(2)) =1

If e_(1) and e_(2) are the eccentricities of the ellipse (x^(2))/(18)+(y^(2))/(4)=1 and the hyperbola (x^(2))/(9)-(y^(2))/(4)=1 respectively and (e_(1), e_(2)) is a point on the ellipse 15x^(2)+3y^(2)=k , then the value of k is equal to

If the eccentricity of the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1and(y^(2))/(b^(2))-(x^(2))/(a^(2))=1" are "e_(1)ande_(2) respectively then prove that : (1)/(e_(1)^(2))+(1)/(e_(2)^(2))=1

If e' is the eccentricity of the ellipse (x^(2))/(a^(2)) + (y^(2))/(b^(2)) =1 (a gt b) , then

If radii of director circle of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 are in the ratio 1:3 and 4e_(1)^(2)-e_(2)^(2)=lambda , where e_1 and e_2 are the eccetricities of ellipse and hyperbola respectively, then the value of lambda is

If e_(1) and e_(2) are eccentricities of the hyperbolas xy=c^(2) and x^2-y^(2)=a^(2) then e_(1)^(2)+e_(2)^(2)=

If e is eccentricity of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (where,a lt b), then

If e_(1) and e_(2) represent the eccentricity of the curves 6x^(2) - 9y^(2) = 144 and 9x^(2) - 16y^(2) = 144 respectively . Then (1)/(e_(1)^(2)) + (1)/(e_(2)^(2)) is equal to