Question

Length of the latus rectum of the parabola $25[(x - 2)^2 + (y - 3)^2] = (3x-4y + 7)^2$ is-

• A. 4
• B. 2
• C. 1/5
• D. 2/5

Answer 1

Answer: (D)

2/5

Answer 2

The given equation is $25[(x - 2)^2 + (y - 3)^2] = (3x-4y + 7)^2$. This can be rewritten as: $5 \sqrt{(x - 2)^2 + (y - 3)^2} = |3x-4y + 7|$ The term $\sqrt{(x - 2)^2 + (y - 3)^2}$ is the distance from a point $(x,y)$ to the point $(2,3)$. The distance from a point $(x,y)$ to the line $3x-4y+7=0$ is given by $d = \frac{|3x-4y+7|}{\sqrt{3^2+(-4)^2}} = \frac{|3x-4y+7|}{5}$. Thus, $|3x-4y+7| = 5d$. Substituting this into the equation, we get: $5 \sqrt{(x - 2)^2 + (y - 3)^2} = 5d$ $\sqrt{(x - 2)^2 + (y - 3)^2} = d$ This equation states that the distance from the point $(x,y)$ to the point $(2,3)$ is equal to the distance from the point $(x,y)$ to the line $3x-4y+7=0$. This is the definition of a parabola where the focus is $(2,3)$ and the directrix is $3x-4y+7=0$.

For a parabola, the distance from the focus to the directrix is $2a$, where $a$ is the distance from the vertex to the focus (or vertex to the directrix). The length of the latus rectum is $4a$.

The distance from the focus $(h,k) = (2,3)$ to the directrix $Ax+By+C=0$ where $A=3, B=-4, C=7$ is: $2a = \frac{|Ah+Bk+C|}{\sqrt{A^2+B^2}} = \frac{|3(2) - 4(3) + 7|}{\sqrt{3^2 + (-4)^2}}$ $2a = \frac{|6 - 12 + 7|}{\sqrt{9 + 16}} = \frac{|1|}{\sqrt{25}} = \frac{1}{5}$

The length of the latus rectum is $4a = 2 \times (2a) = 2 \times \frac{1}{5} = \frac{2}{5}$.