The point on the parabola y^2 = 8x at which t | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the normal to the parabola $y^2 = 4ax$ at point $(x_1, y_1

Vedclass · Accepted Answer

$(6, -4\sqrt{3})$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the normal to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $m = -\frac{y_1}{2a}$.The normal is inclined at $60^\circ$ to the $x$-axis,so its slope $m = 	an(60^\circ) = \sqrt{3}$.Equating the slopes: $\sqrt{3} = -\frac{y_1}{2(2)} = -\frac{y_1}{4}$.Thus,$y_1 = -4\sqrt{3}$.Since the point lies on the parabola $y^2 = 8x$,we have $(-4\sqrt{3})^2 = 8x_1$.$16 	imes 3 = 8x_1 \implies 48 = 8x_1 \implies x_1 = 6$.Therefore,the required point is $(6, -4\sqrt{3})$.

The point on the parabola $y^2 = 8x$ at which the normal is inclined at $60^\circ$ to the $x$-axis has the coordinates:

Similar Questions

What is the locus of the vertices $(x, y)$ of the parabolas $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$?

The point of intersection of tangents at the ends of the latus rectum of the parabola $y^2 = 4x$ is

Find the equation of the latus rectum of the parabola $x^2 = -12y$.

If $(at^2, 2at)$ are the coordinates of one end of a focal chord of the parabola $y^2 = 4ax$,find the coordinates of the other end.

Find the equation of the parabola with vertex $(-3, 0)$ and directrix $x + 5 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module