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 $a = 4$.Let the coordinates of the ends of the fo

Vedclass · Accepted Answer

$(16, 16)$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.Let the coordinates of the ends of the focal chord be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$.For a focal chord,the relation between the parameters is $t_1 t_2 = -1$.Given one end is $(1, -4)$,we have $2at_1 = -4$.Substituting $a = 4$,we get $2(4)t_1 = -4$,which gives $t_1 = -1/2$.Since $t_1 t_2 = -1$,we have $t_2 = -1 / t_1 = -1 / (-1/2) = 2$.The other end is $(at_2^2, 2at_2) = (4(2)^2, 2(4)(2)) = (4 	imes 4, 16) = (16, 16)$.

If one end of a focal chord of the parabola $y^2 = 16x$ is $(1, -4)$,find the other end.

Similar Questions

The angle between the tangents drawn from the origin to the parabola $y^2 = 4a(x - a)$ is ............... $^\circ$.

If the normals at two points on the parabola $y^2 = 4ax$ meet on the parabola,then what is the product of the ordinates of these two points?

Let the point $P$ of the focal chord $PQ$ of the parabola $y^2=16x$ be $(1, -4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$,where $\operatorname{gcd}(m, n)=1$,then $m^2+n^2$ is equal to:

Find the length of the chord intercepted by the line $4y = 3x - 48$ on the parabola $y^{2} = 64x$.

If the normal at the point $t_1$ (i.e.,at $(at_1^2, 2at_1)$) on the parabola $y^2 = 4ax$ meets the parabola again at the point $t_2$,then $t_1t_2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module