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. $(3x+4y-4{)}^2=10(4x-3y+7)$
• B. $(3x+4y-4{)}^2=4(4x-3y+7)$
• C. $(3x+4y-4{)}^2=20(4x-3y+7)$
• D. $(3x+4y-4{)}^2=5(4x-3y+7)$

Answer 1

Answer: (C)

$(3x+4y-4{)}^2=20(4x-3y+7)$

Since tangent at vertex is perpendicular to axis of parabola
So we can take $y^{\prime }$ axis as tangent at vertex and axis as $x^{\prime }$ axis
Its like $(x^{\prime },y^{\prime })$ frame is shifted and rotated frame of $(x,y)$
$y^{\prime 2}=4a{x}^{\prime }$ where 4a=length of latus rectum
$⟹y^{\prime 2}=4x^{\prime }$
$y^{\prime }$ is the distance from $x^{\prime }$ axis and vice versa
So $y^{\prime }=\frac{3x+4y-4}{5}$ and $x^{\prime }=\frac{4x-3y+7}{5}$
Apply this above on equation parabola
$(3x+4y-4{)}^2=20(4x-3y+7)$