If one end of a focal chord of the parabola y | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The focus of the parabola is $S(a, 0) = (

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The focus of the parabola is $S(a, 0) = (4, 0)$.Let the coordinates of one end of the focal chord be $A(at_1^2, 2at_1) = (1, 4)$.Since $a = 4$,we have $4t_1^2 = 1$ $\Rightarrow t_1^2 = \frac{1}{4}$ $\Rightarrow t_1 = \frac{1}{2}$ (since $y > 0$).For a focal chord,the product of the parameters of the endpoints is $t_1 t_2 = -1$. Thus,$t_2 = -\frac{1}{t_1} = -2$.The length of a focal chord with parameter $t$ is given by $L = a(t + \frac{1}{t})^2$.Substituting $t = t_1 = \frac{1}{2}$,we get:$L = 4(\frac{1}{2} + \frac{1}{1/2})^2 = 4(\frac{1}{2} + 2)^2 = 4(\frac{5}{2})^2 = 4 	imes \frac{25}{4} = 25$.

If one end of a focal chord of the parabola $y^2 = 16x$ is at $(1, 4)$,then the length of this focal chord is

Similar Questions

If the line $y=2x+k$ is a normal to the parabola $y^2=4x$,then $k=$

The area of the triangle formed by the lines joining the vertex of the parabola $x^2=12y$ to the ends of its latus rectum is equal to $...$ sq. units.

What is the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$?

Find the coordinates of the focus,axis,the equation of the directrix,and the length of the latus rectum of the parabola $y^{2}=8x$.

Let the normal at the point $P$ on the parabola $y^{2} = 6x$ pass through the point $(5, -8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$,then the ordinate of the point $Q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module