Question

Consider the parabola \(25[(x-2)^2 + (y+5)^2] = (3x+4y-1)^2\), match the characteristic of this parabola given in List-I with its corresponding item in List-II. 

• A. I-B, II-E, III-C, IV-D
• B. I-D, II-A, III-C, IV-B
• C. I-B, II-A, III-C, IV-D
• D. I-D, II-B, III-C, IV-A

Hint:

For parabolas represented in complex forms, always attempt to simplify to a recognizable conic section form (vertex form or standard form) to more easily identify geometric features.

Answer 1

The Correct Option is A

Let the given parabola be $$25[(x-2)^2 + (y+5)^2] = (3x+4y-1)^2$$ We can rewrite this as $$(x-2)^2 + (y+5)^2 = \left(\frac{3x+4y-1}{5}\right)^2$$ This is in the form $SP^2 = PM^2$, where $S(2, -5)$ is the focus and $PM$ is the perpendicular distance from $(x,y)$ to the directrix $3x+4y-1=0$. Therefore, the focus is $S(2, -5)$ and the directrix is $3x+4y-1=0$. I. Vertex: The vertex is the midpoint of the perpendicular distance from the focus to the directrix. Let $V(x_v, y_v)$ be the vertex. The line perpendicular to the directrix and passing through the focus is $$\frac{x-2}{3} = \frac{y+5}{4} = \lambda$$ So, $x = 3\lambda + 2$ and $y = 4\lambda - 5$. The foot of the perpendicular from the focus to the directrix is obtained by substituting $x$ and $y$ into the equation of the directrix: $$3(3\lambda + 2) + 4(4\lambda - 5) - 1 = 0$$ $$9\lambda + 6 + 16\lambda - 20 - 1 = 0$$ $$25\lambda - 15 = 0$$ $$\lambda = \frac{15}{25} = \frac{3}{5}$$ The foot of the perpendicular is $$x = 3\left(\frac{3}{5}\right) + 2 = \frac{9}{5} + 2 = \frac{19}{5}$$ $$y = 4\left(\frac{3}{5}\right) - 5 = \frac{12}{5} - 5 = \frac{12-25}{5} = -\frac{13}{5}$$ The foot of the perpendicular is $\left(\frac{19}{5}, -\frac{13}{5}\right)$. The vertex is the midpoint of the focus and the foot of the perpendicular: $$x_v = \frac{2 + \frac{19}{5}}{2} = \frac{\frac{10+19}{5}}{2} = \frac{29}{10}$$ $$y_v = \frac{-5 - \frac{13}{5}}{2} = \frac{\frac{-25-13}{5}}{2} = \frac{-38}{10}$$ Thus, the vertex is $V\left(\frac{29}{10}, -\frac{38}{10}\right)$. So, I-B. II. Length of latus rectum: The length of the latus rectum is twice the distance between the focus and the directrix. $$LR = 2 \times \frac{|3(2) + 4(-5) - 1|}{\sqrt{3^2 + 4^2}} = 2 \times \frac{|6 - 20 - 1|}{5} = 2 \times \frac{15}{5} = 2 \times 3 = 6$$ So, II-E. III. Directrix:
The directrix is $3x+4y-1=0$. So, III-C. IV. One end of the latus rectum:
The latus rectum passes through the focus and is perpendicular to the axis of the parabola.
The axis of the parabola is the line joining the focus and the vertex.
The latus rectum is parallel to the directrix. The length of the latus rectum is 6.
The distance from the focus to the end of the latus rectum is 3.
The slope of the directrix is $-\frac{3}{4}$.
The slope of the latus rectum is $-\frac{3}{4}$.
The equation of the latus rectum is $3x+4y+c=0$.
Since it passes through $(2, -5)$, we have
$3(2) + 4(-5) + c = 0$
$6 - 20 + c = 0$
$c = 14$
So, the latus rectum is $3x+4y+14=0$.
The distance from the focus $(2, -5)$ to the end of the latus rectum is 3.
Let $(x, y)$ be one end of the latus rectum.
$(x-2)^2 + (y+5)^2 = 3^2 = 9$
Also, $3x+4y+14=0$.
$3x = -4y-14$
$x = \frac{-4y-14}{3}$
Substituting in the distance equation:
$\left(\frac{-4y-14}{3} - 2\right)^2 + (y+5)^2 = 9$
$\left(\frac{-4y-14-6}{3}\right)^2 + (y+5)^2 = 9$
$\left(\frac{-4y-20}{3}\right)^2 + (y+5)^2 = 9$
$\frac{16(y+5)^2}{9} + (y+5)^2 = 9$
$(y+5)^2\left(\frac{16}{9} + 1\right) = 9$
$(y+5)^2\left(\frac{25}{9}\right) = 9$
$(y+5)^2 = \frac{81}{25}$
$y+5 = \pm \frac{9}{5}$
$y = -5 \pm \frac{9}{5}$
$y_1 = -5 + \frac{9}{5} = \frac{-25+9}{5} = -\frac{16}{5}$
$y_2 = -5 - \frac{9}{5} = \frac{-25-9}{5} = -\frac{34}{5}$
If $y = -\frac{16}{5}$, $x = \frac{-4(-\frac{16}{5}) - 14}{3} = \frac{\frac{64}{5} - \frac{70}{5}}{3} = \frac{-6}{15} = -\frac{2}{5}$
If $y = -\frac{34}{5}$, $x = \frac{-4(-\frac{34}{5}) - 14}{3} = \frac{\frac{136}{5} - \frac{70}{5}}{3} = \frac{66}{15} = \frac{22}{5}$
One end is $\left(-\frac{2}{5}, -\frac{16}{5}\right)$. So, IV-D. Final Answer: The final answer is $\boxed{(1)}$