Find the coordinates of the point of intersec | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1

Vedclass · Accepted Answer

$(-1, 0)$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1, 2)$ and $(1, -2)$.The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.For the point $(1, 2)$,the tangent is $y(2) = 2(1)(x + 1) \implies y = x + 1$.For the point $(1, -2)$,the tangent is $y(-2) = 2(1)(x + 1) \implies -y = x + 1$.Adding the two equations: $0 = 2(x + 1) \implies x = -1$.Substituting $x = -1$ into $y = x + 1$,we get $y = -1 + 1 = 0$.Thus,the intersection point is $(-1, 0)$.

Find the coordinates of the point of intersection of the tangents at the endpoints of the latus rectum of the parabola $y^2 = 4x$.

Similar Questions

$A$ parabola having its axis parallel to the $Y$-axis passes through the points $(0, 2/5)$,$(4, -2)$,and $(1, 8/5)$. Which of the following points lies on this parabola?

If ${x^2} + 6x + 20y - 51 = 0$,then the axis of the parabola is

The value of $\lambda$ for which the curve $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$ represents a parabola is

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

If the parabolas $y^2 = 4b(x - c)$ and $y^2 = 8ax$ have a common normal,then which one of the following is a valid choice for the ordered triad $(a, b, c)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module