Let an ellipse with centre (1,0) and latus re | vedclass

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

Question

(A) The equation of the ellipse is $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{1}{2}$,which

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The equation of the ellipse is $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{1}{2}$,which implies $4b^2 = a$ (Equation $1$).The minor axis endpoints are $(1, b)$ and $(1, -b)$,and the foci are $(1 \pm ae, 0)$. The angle subtended by the minor axis at a focus is $60^{\circ}$.Considering the triangle formed by the focus $(1+ae, 0)$ and the endpoints of the minor axis $(1, b)$ and $(1, -b)$,the angle at the focus is $60^{\circ}$,so the half-angle is $30^{\circ}$.Thus,$	an(30^{\circ}) = \frac{b}{ae} = \frac{1}{\sqrt{3}}$,which implies $ae = b\sqrt{3}$.Squaring both sides,$a^2e^2 = 3b^2$. Since $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 3b^2$,so $a^2 = 4b^2$ (Equation $2$).From Equation $1$,$b^2 = \frac{a}{4}$. Substituting this into Equation $2$,$a^2 = 4(\frac{a}{4}) = a$.Since $a > 0$,we get $a = 1$. Then $b^2 = \frac{1}{4}$,so $b = \frac{1}{2}$.The length of the major axis is $2a = 2(1) = 2$,and the length of the minor axis is $2b = 2(\frac{1}{2}) = 1$.The square of the sum of the lengths is $(2a + 2b)^2 = (2 + 1)^2 = 3^2 = 9$.

List-$I$	List-$II$
$(i)$ The equation of the major axis	$(p)$ $3x = 34$
$(ii)$ The equation of a directrix	$(q)$ $y = 2$
$(iii)$ The equation of a latus rectum	$(r)$ $x + y = 9$
	$(s)$ $x = 6$
	$(t)$ $x = 3$
	$(u)$ $3y = 34$

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

Similar Questions

The eccentricity of the ellipse $\left( \frac{x - 3}{y} \right)^2 + \left( 1 - \frac{4}{y} \right)^2 = \frac{1}{9}$ is

If $\theta$ and $\phi$ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then $\theta - \phi$ is equal to

Let $L$ be a common tangent line to the curves $4x^{2} + 9y^{2} = 36$ and $(2x)^{2} + (2y)^{2} = 31$. Then the square of the slope of the line $L$ is ..... .

The number of values of $c$ such that the line $y=4x+c$ touches the curve $\frac{x^{2}}{4}+y^{2}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module