If the maximum distance of the normal to the | vedclass

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

Question

(B) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $ax \sec 	heta - b

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $ax \sec 	heta - by \operatorname{cosec} 	heta = a^2 - b^2$.Here $a^2 = 4$,so the equation is $2x \sec 	heta - by \operatorname{cosec} 	heta = 4 - b^2$.The perpendicular distance $d$ from the origin $(0,0)$ is $d = \frac{|4 - b^2|}{\sqrt{4 \sec^2 	heta + b^2 \operatorname{cosec}^2 	heta}}$.To maximize $d$,we minimize $f(	heta) = 4 \sec^2 	heta + b^2 \operatorname{cosec}^2 	heta = 4(1 + 	an^2 	heta) + b^2(1 + \cot^2 	heta) = (4 + b^2) + 4 	an^2 	heta + b^2 \cot^2 	heta$.Using $AM$-$GM$ inequality,$4 	an^2 	heta + b^2 \cot^2 	heta \geq 2 \sqrt{4 	an^2 	heta \cdot b^2 \cot^2 	heta} = 4b$.The minimum value is $(4 + b^2 + 4b) = (2 + b)^2$.Thus,$d_{max} = \frac{4 - b^2}{\sqrt{(2 + b)^2}} = \frac{(2 - b)(2 + b)}{2 + b} = 2 - b$.Given $d_{max} = 1$,we have $2 - b = 1$,so $b = 1$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the maximum distance of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$,where $b < 2$,from the origin is $1$,then the eccentricity of the ellipse is:

Similar Questions

$P$ is a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. When the area of $\Delta PSS'$ is maximum,the inradius of $\Delta PSS'$ ($S$ and $S'$ are foci) is equal to:

If $x+\sqrt{3} y=3$ is the tangent to the ellipse $2 x^2+3 y^2=k$ at a point $P$,then the equation of the normal to this ellipse at $P$ is

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

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$. If one of its directrices is $x = -4$,then the equation of the normal to it at $\left(1, \frac{3}{2}\right)$ is

The eccentric angle of a point on the ellipse $x^2 + 3y^2 = 6$ at a distance of $2$ units from the centre of the ellipse is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module