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 $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$. Dividing by $20$,we get $(x-3)^2 + (y-1)^2 = \left(\frac{4x-2y-5}{\sqr

Vedclass · Accepted Answer

$\sqrt{5}$

Vedclass · Answer

(C) The given equation of the parabola is $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$. Dividing by $20$,we get $(x-3)^2 + (y-1)^2 = \left(\frac{4x-2y-5}{\sqrt{20}}ight)^2$. This is in the form $SP^2 = PM^2$,where $S(3,1)$ is the focus and $4x-2y-5=0$ is the directrix. The distance from the focus to the directrix is $2a = \frac{|4(3)-2(1)-5|}{\sqrt{4^2+(-2)^2}} = \frac{|12-2-5|}{\sqrt{16+4}} = \frac{5}{\sqrt{20}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$. The length of the latus rectum is $4a = 2(2a) = 2 	imes \frac{\sqrt{5}}{2} = \sqrt{5}$.

The length of the latus rectum of the parabola $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$ is

Similar Questions

The angle between tangents to the parabola $y^2 = 4ax$ at the points where it intersects with the line $x - y - a = 0$ is

If a focal chord of the parabola $y^2 = 4x$ makes an angle $45^{\circ}$ with the positive $X$-axis,then the slopes of the normals drawn at the ends of the focal chord will satisfy the equation:

Find the locus of the midpoint of the chord of the parabola $y^2 = 4x$ drawn from the vertex.

What is the angle subtended by the latus rectum of the parabola $y^2 = ax$ at its vertex?

If the tangent to the curve $y^2 = 4x$ at point $(1, 2)$ cuts the coordinate axes at points $A$ and $B$,then the area of $\Delta AOB$ is (where $'O'$ is the origin).

Vedclass Test Series

Exam Paper Generator

Online Exam Module