The length of the latus rectum of the parabol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the parabola is $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$. Dividing by $169$,we get: $(x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{

Vedclass · Accepted Answer

$\frac{28}{13}$

Vedclass · Answer

(C) The given equation of the parabola is $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$. Dividing by $169$,we get: $(x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{13}ight)^2$. This is in the form $SP^2 = PM^2$,where $S$ is the focus $(1, 3)$ and $PM$ is the perpendicular distance from point $P(x, y)$ to the directrix $5x-12y+17=0$. The distance between the focus and the directrix is $2a$. $2a = \left|\frac{5(1)-12(3)+17}{\sqrt{5^2+(-12)^2}}ight| = \left|\frac{5-36+17}{13}ight| = \left|\frac{-14}{13}ight| = \frac{14}{13}$. The length of the latus rectum is $4a$. Since $2a = \frac{14}{13}$,then $4a = 2 	imes \frac{14}{13} = \frac{28}{13}$.

The length of the latus rectum of the parabola $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$ is

Similar Questions

Let $P, Q,$ and $R$ be three co-normal points on the parabola $y^2 = 4ax$. Then the correct statement$(s)$ is/are:

The length of the latus rectum of the parabola $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$ is

$A$ particle is moving in the $xy$-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the $x$-axis at $Q$. If the $y$-axis bisects the segment $PQ$,then $C$ is a parabola with

Tangents drawn from the point $(-8, 0)$ to the parabola $y^2 = 8x$ touch the parabola at $P$ and $Q$. If $F$ is the focus of the parabola,then the area of the triangle $PFQ$ (in sq. units) is equal to

What is the length of the latus rectum of the parabola $x = ay^2 + by + c$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module