Question:
The length of the latus rectum of the parabola $bx^2 - 4ay + dx + e = 0$ is
The length of the latus rectum of the parabola $bx^2 - 4ay + dx + e = 0$ is
Updated On: May 12, 2024
- $ \frac{a}{b} $
- $4a$
- $ \frac{4b}{a} $
- $ \frac{4a}{b} $
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The Correct Option is D
Solution and Explanation
$bx^2 - 4ay + dx + e = 0$
$\Rightarrow x^{2} +\frac{dx}{b} + \frac{d^{2}}{4b^{2}} - \frac{d^{2}}{4b^{2}} = \frac{4ay}{b} - \frac{e}{b} $
$ \Rightarrow \left(x+ \frac{d}{2b}\right)^{2} = \frac{4ay}{b} + \frac{d^{2}}{4b^{2}} - \frac{e}{b} $
$= \frac{4a}{b} \left[y + \frac{d^{2}}{16 \,ab} - \frac{e}{4a}\right]$
On comparing with general equation of parabola $(X + x)^2 = 4a(Y + y)$, we have
Length of latus rectum $ = \frac{4a}{b} $
$\Rightarrow x^{2} +\frac{dx}{b} + \frac{d^{2}}{4b^{2}} - \frac{d^{2}}{4b^{2}} = \frac{4ay}{b} - \frac{e}{b} $
$ \Rightarrow \left(x+ \frac{d}{2b}\right)^{2} = \frac{4ay}{b} + \frac{d^{2}}{4b^{2}} - \frac{e}{b} $
$= \frac{4a}{b} \left[y + \frac{d^{2}}{16 \,ab} - \frac{e}{4a}\right]$
On comparing with general equation of parabola $(X + x)^2 = 4a(Y + y)$, we have
Length of latus rectum $ = \frac{4a}{b} $
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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.