If a neq 0 and the line 2bx + 3cy + 4d = 0 pa | vedclass

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

Question

(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$. Therefore,the line $y = x$ passes through their points o

Vedclass · Accepted Answer

$d^2 + (2b + 3c)^2 = 0$

Vedclass · Answer

(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$. Therefore,the line $y = x$ passes through their points of intersection. Solving $y^2 = 4ax$ and $y = x$: $x^2 = 4ax \Rightarrow x(x - 4a) = 0 \Rightarrow x = 0$ or $x = 4a$. When $x = 0$,$y = 0$. When $x = 4a$,$y = 4a$. Thus,the points of intersection are $(0, 0)$ and $(4a, 4a)$. Since the line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$: $2b(0) + 3c(0) + 4d = 0 \Rightarrow 4d = 0 \Rightarrow d = 0$. Since it also passes through $(4a, 4a)$: $2b(4a) + 3c(4a) + 4d = 0 \Rightarrow 8ab + 12ac + 4d = 0$. Since $d = 0$ and $a 
eq 0$,we have $8ab + 12ac = 0 \Rightarrow 4a(2b + 3c) = 0 \Rightarrow 2b + 3c = 0$. Therefore,$d^2 + (2b + 3c)^2 = 0^2 + 0^2 = 0$.

If $a \neq 0$ and the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of the parabolas $y^2 = 4ax$ and $x^2 = 4ay$,then

Similar Questions

The length of the latus rectum of a parabola,whose vertex and focus are on the positive $x$-axis at a distance $R$ and $S$ $(S > R)$ respectively from the origin,is:

If one of the vertices of an equilateral triangle inscribed in the parabola $y^2=12x$ coincides with the vertex of the parabola, then the area (in sq. units) of that triangle is (in $\sqrt{3}$)

The length of the latus rectum of the parabola ${y^2} = 5x + 4y + 1$ is

The line $y=x+1$ is a tangent to the curve $y^{2}=4x$ at the point

Find the equation of the parabola that satisfies the following conditions: Focus $(0, -3)$,directrix $y = 3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module