For an ellipse with eccentricity e = frac{1}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that,eccentricity $e = \frac{1}{2}$ and the equation of the directrix is $\frac{a}{e} = 4$. Since $e = \frac{1}{2}$,we have $\frac{a}{1/2} =

Vedclass · Accepted Answer

$3x^2 + 4y^2 = 12$

Vedclass · Answer

(B) Given that,eccentricity $e = \frac{1}{2}$ and the equation of the directrix is $\frac{a}{e} = 4$. Since $e = \frac{1}{2}$,we have $\frac{a}{1/2} = 4$,which implies $a = 2$. Using the relation $b^2 = a^2(1 - e^2)$,we get: $b^2 = 4(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$. Therefore,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Substituting the values,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one directrix is $x = 4$,then the equation of the ellipse is

Similar Questions

Which plant group exhibits a haplodiplontic life cycle pattern?

Statement $S$: Presence of a rigid cell wall and nucleoprotein as genetic material.
Statement $R$: Examples include $Euglena$,$Dinoflagellates$,$Amoeba$,and $Paramecium$.

The product of the lengths of the perpendiculars from any point on the hyperbola $x^2 - y^2 = 8$ to its asymptotes is:

Which one of the following ions is the most stable in aqueous solution?
$(At. \ No. \ Ti = 22, V = 23, Cr = 24, Mn = 25)$

$A$ straight conductor of length $4 ~m$ moves at a speed of $10 ~m/s$. When the conductor makes an angle of $30^{\circ}$ with the direction of the magnetic field of induction $0.1 ~Wb/m^2$,the induced emf is: (in $~V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module