Given that, the ends of the major axis\( (0, ±√5)\) , ends of minor axis \(( ±1, 0)\)
Here, the major axis is along the \(y-axis.\)
Therefore, the equation of the ellipse will be of the form \(\dfrac{x^2}{b^2}+ \dfrac{y^2}{a^2} = 1\), where ‘a’ is the semi-major axis.
Accordingly, a = √5 and b = 1.
∴ The equation for the ellipse \(\dfrac{x^2}{1^2}+ \dfrac{y^2}{(√5)^2} = 1\)or \(\dfrac{x^2}{1}+ \dfrac{y^2}{5} = 1\) (Ans)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Read More: Conic Section
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}