CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

Solution

The correct option is B $$(3x+4y-4)^{2}=20(4x-3y+7)$$
Since tangent at vertex is perpendicular to axis of parabola
So we can take $$y'$$ axis as tangent at vertex and axis as $$x'$$ axis
Its 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 versa
So $$y'=\dfrac{3x+4y-4}{5}$$ and $$x'=\dfrac{4x-3y+7}{5}$$
Apply this above on equation parabola
$$(3x+4y-4)^{2}=20(4x-3y+7)$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image