The angle between the two curves y^2 = 4(x + | vedclass

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

Question

(B) The given equations are $y^2 = 4(x + 1)$ and $x^2 = 4(y + 1)$.These represent two parabolas.The first parabola $y^2 = 4(x + 1)$ has its vertex at

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) The given equations are $y^2 = 4(x + 1)$ and $x^2 = 4(y + 1)$.These represent two parabolas.The first parabola $y^2 = 4(x + 1)$ has its vertex at $(-1, 0)$ and opens to the right along the $x$-axis.The second parabola $x^2 = 4(y + 1)$ has its vertex at $(0, -1)$ and opens upwards along the $y$-axis.Since the axis of the first parabola is the $x$-axis and the axis of the second parabola is the $y$-axis,and these axes are perpendicular to each other,the angle of intersection between these two curves at their point of intersection is $90^\circ$.

The angle between the two curves $y^2 = 4(x + 1)$ and $x^2 = 4(y + 1)$ is .............. $^\circ$.

Similar Questions

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2}=10x$.

Two tangents are drawn from the point $(-1, -2)$ to the parabola $y^2 = 4x$. If $\theta$ is the angle between these tangents,then $\tan \theta = $

What is the equation of the chord of contact drawn from the point $(2, 4)$ to the parabola $y^2 = 4x$?

If the equation of a system of parallel chords of the parabola $y^2 = \frac{2}{3}x$ is $y + 2x + 1 = 0$,find its diameter.

The point on the parabola $y^{2}=64x$ which is nearest to the line $4x+3y+35=0$ has coordinates

Vedclass Test Series

Exam Paper Generator

Online Exam Module