Question:
The equation of a directrix of the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1 $ is
The equation of a directrix of the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1 $ is
Updated On: May 12, 2022
- 3y = 5
- y = 5
- 3y = 25
- y = 3
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The Correct Option is C
Solution and Explanation
Equation of ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 $ , where a > b
Given, $ \frac{x^{2}}{16} + \frac{y^{2}}{75} = 1 \Rightarrow b = 4 , a=5$
But $ e=\sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 -\frac{16}{25}} $
$ \Rightarrow e = \frac{3}{5} $
$ \therefore $ equation of directrix $ y = \pm \frac{a}{e} $
$ \therefore y = \pm \frac{5}{\frac{3}{5}} \Rightarrow 3y = \pm 25 $
Given, $ \frac{x^{2}}{16} + \frac{y^{2}}{75} = 1 \Rightarrow b = 4 , a=5$
But $ e=\sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 -\frac{16}{25}} $
$ \Rightarrow e = \frac{3}{5} $
$ \therefore $ equation of directrix $ y = \pm \frac{a}{e} $
$ \therefore y = \pm \frac{5}{\frac{3}{5}} \Rightarrow 3y = \pm 25 $
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.