Question

Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3,3) and directrix is 3x-4y=2.Find also the length of the latus-rectum.

Answer 1

The axis of the parabola is a line $\perp$ to the directrix and passing through focus.The equation of a line $\perp$ to 3x-4y-2=0 is

$y=\frac{-4}{3}+\lambda \left[\begin{array}{cc}& \because m_1{m}_2=-1\\ & \Rightarrow m_2=\frac{-1}{m_1}andy=m_{2}x+\lambda \end{array}\right]\Rightarrow 3x+4y=3\lambda \text{This will pass through focus (3,3) if,}3\times 3+4\times 3=3\lambda \Rightarrow 9+12=3\lambda \Rightarrow 21=3\lambda \Rightarrow \lambda =\frac{21}{3}=7so,theequationofaxisof3y+4x=3\times 7=21\Rightarrow 3y+4x=21...\left(i\right)\text{And the equation of directrix is}\left(ii\right)by3weget3x-4y=2...\left(ii\right)\text{Multiplying equation (i) by 4 and equation (ii) by 3,we~ get}16x+12y=84...\left(iii\right)9x-12y=6...\left(iv\right)\text{Adding equation (iii) and (iv),we get 16x+9x=84+6}\Rightarrow 25x=90\Rightarrow x=\frac{90}{25}=\frac{18}{5}\text{Putting x=}\frac{18}{5}inequation\left(i\right),weget3y+4\times \frac{18}{5}=21\Rightarrow 3y+\frac{72}{5}=21\Rightarrow 3y=21-\frac{72}{5}\Rightarrow 3y=\frac{105-72}{5}\Rightarrow 3y=\frac{33}{5}\Rightarrow y=\frac{11}{5}\text{Hence,the required point of intersection is}(\frac{18}{5},\frac{11}{5}).\text{Also,length of the latus rectum=2(Length of the perpendicular from the focus on the directrix)}=2\left|\frac{3\left(3\right)+(-4)3-2}{\sqrt{16+9}}\right|=2\left|\frac{-5}{\sqrt{16+9}}\right|=2units.$