Question

Find the equation of the parabola whose latus rectum is 4units and axis is the line 3x+4y-4=0 and tangent at vertex is 4x-3y+7=0

Answer 1

Please see the attachment

Answer 2

The equation of parabola with the given conditions  is (3x+4y-4)^2=+20\times (4x-3y+7) or (3x+4y-4)^2=-20\times (4x-3y+7).

 

Step-by-step explanation:

Given:

Latus rectum = 4 units and axis is the line 3x+4y-4=0 and tangent at vertex is 4x-3y+7=0

Now using distance formula,

$length of PN = |\frac{Ax_1+By_1+C}{A^2+B^2} |$  

here$(x_1, y_1)$  is the point and the line is Ax+By+C=0.

length of PN =$|\frac{3x+4y-4}{\sqrt{3^2+4^2} } |$

Squaring both sides, we get

$PN^2=(\frac{3x+4y-4}{\sqrt{3^2+4^2} } )^2$

               =$\frac{(3x+4y-4)^2}{3^2+4^2}$

$PN^2=\frac{(3x+4y-4)^2}{25}$         (1)

Now PM = $|\frac{Ax_1+By_1+C}{A^2+B^2} |$

               = $|\frac{4x-3y+7}{\sqrt{3^2+4^2} } |$

               = $|\frac{4x-3y+7}{\sqrt{25} } |$

               =  $|\frac{4x-3y+7}{5} } |$

Now equation of parabola will be

$PN^2$  = length of latus rectum \times PM

$\frac{(3x+4y-4)^2}{25} = 4 \times |\frac{4x-3y+7}{5} } |$   [from equation (1)]

${(3x+4y-4)^2}= 25\times 4\times|\frac{4x-3y+7}{5} |$

${(3x+4y-4)^2}= 5\times 4\times|{4x-3y+7 |$

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

Here mode determines that parabola can be either be on positive X axis or negative X axis.

Therefore the vertex 4x-3y+7=0 can be positive or negative.

Hence the equation of parabola with the given conditions  is (3x+4y-4)^2=+20\times (4x-3y+7) or (3x+4y-4)^2=-20\times (4x-3y+7).