The two parabolas y^2 = 4x and x^2 = 4y inter | vedclass

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

Question

(C) Solving $x^2 = 4y$ and $y^2 = 4x$,we get $x = 0, y = 0$ and $x = 4, y = 4$.Since the abscissa is not zero,the point $P$ is $(4, 4)$.The equation o

Vedclass · Accepted Answer

The tangents to each curve at $P$ make complementary angles with the $x$-axis

Vedclass · Answer

(C) Solving $x^2 = 4y$ and $y^2 = 4x$,we get $x = 0, y = 0$ and $x = 4, y = 4$.Since the abscissa is not zero,the point $P$ is $(4, 4)$.The equation of the tangent to $y^2 = 4x$ at $(4, 4)$ is $y(4) = 2(x + 4)$,which simplifies to $2x - y + 4 = 0$ (Slope $m_1 = 2$).The equation of the tangent to $x^2 = 4y$ at $(4, 4)$ is $x(4) = 2(y + 4)$,which simplifies to $x - 2y + 4 = 0$ (Slope $m_2 = 1/2$).Let $	heta_1$ and $	heta_2$ be the angles the tangents make with the $x$-axis. Then $	an 	heta_1 = 2$ and $	an 	heta_2 = 1/2$.Since $	an 	heta_1 = \cot 	heta_2 = 	an(90^\circ - 	heta_2)$,we have $	heta_1 = 90^\circ - 	heta_2$,or $	heta_1 + 	heta_2 = 90^\circ$.Thus,the tangents make complementary angles with the $x$-axis.

The two parabolas $y^2 = 4x$ and $x^2 = 4y$ intersect at a point $P$,whose abscissa is not zero,such that

Similar Questions

Let a focal chord $12x + 5y - 27 = 0$ of the parabola $y^2 = kx$ intersect the parabola at the points $P$ and $P^{\prime}$. If $S$ is the focus of this parabola,then $9(SP + SP^{\prime}) = $

The Cartesian equation of the parabola $x = -2 + 2t^2$,$y = 2 + 4t$ is

If $y^2=16x$ is the given parabola,then the point of intersection of the focal chord passing through the point $(2,2)$ and the double ordinate of length $24$ is

The equation of the common tangent touching the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $x$-axis is

If $y = 2x - 3$ is a tangent to the parabola $y^2 = 4a(x - \frac{1}{3})$,then $a$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module