Question

# Equation of the parabola with axis  $$3x+4y-4=0$$, the tangent at the vertex $$4x-3y+7=0$$, and with length of latus rectum $$4$$ is

A
(3x+4y4)2=10(4x3y+7)
B
(3x+4y4)2=4(4x3y+7)
C
(3x+4y4)2=20(4x3y+7)
D
(3x+4y4)2=5(4x3y+7)

Solution

## The correct option is B $$(3x+4y-4)^{2}=20(4x-3y+7)$$Since tangent at vertex is perpendicular to axis of parabolaSo we can take $$y'$$ axis as tangent at vertex and axis as $$x'$$ axisIts like $$(x',y')$$ frame is shifted and rotated frame of $$(x,y)$$$$y'^{2}=4ax'$$ where 4a=length of latus rectum$$\implies y'^{2}=4x'$$$$y'$$ is the distance from $$x'$$ axis and vice versaSo $$y'=\dfrac{3x+4y-4}{5}$$ and $$x'=\dfrac{4x-3y+7}{5}$$Apply this above on equation parabola $$(3x+4y-4)^{2}=20(4x-3y+7)$$Mathematics

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