The length of the latus rectum of the conic 2 | vedclass

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

Question

(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$. Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$. This is in

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$. Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}ight)^2$. This is in the form $SP^2 = PM^2$,where $S(2,3)$ is the focus and $3x-4y+7=0$ is the directrix. Here,the distance from the focus to the directrix is $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}ight| = \left|\frac{6-12+7}{5}ight| = \frac{1}{5}$. The length of the latus rectum of a parabola is $4a$. Therefore,length $= 4 	imes \frac{1}{5} = \frac{4}{5}$.

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

Similar Questions

Find the equation of the tangent to the parabola $x^2 = y$ at one of the endpoints of its latus rectum in the first quadrant.

Let $Z$ be the point of intersection of the axis and the directrix of the parabola $4x^2 - 12x + 4y + 5 = 0$. If $S$ is its focus,then the point which divides $SZ$ in the ratio $2:1$ is

The length of the latus rectum of the parabola whose focus is at $(1, -2)$ and directrix is the line $x + y + 3 = 0$ is

If the angle between the tangents drawn through the point $(-2, -1)$ to the parabola $y^2 = 4x$ is $\theta$,then $\tan 2\theta =$

Find the parametric coordinates of any point on the parabola whose focus is $(0, 1)$ and directrix is $x + 2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module