Determining different elements of parabola
If the parabola y2=4 ax{y^2} = 4\,\,axy2=4ax passes through the point (−3,2),\left( { - 3,2} \right),(−3,2), then the length of its latus rectum is-
2 / 3
4 / 3
1 / 3
4
y2=4ax(−3,2)→4=4×ax−3∴a=−1/3 \begin{array}{l} y^{2}=4 a x \\ (-3,2) \rightarrow 4=4 \times a x-3 \\ \therefore a=-1 / 3 \end{array} y2=4ax(−3,2)→4=4×ax−3∴a=−1/3
∴ \therefore ∴ latus rectum ∣4a∣=43 |4 a|=\frac{4}{3} ∣4a∣=34
Angle between the parabola y2=4b(x−2a+b)y^{2} = 4b(x - 2a + b)y2=4b(x−2a+b) and x2+4a(y−2b−a)=0x^{2} + 4a(y - 2b - a) = 0x2+4a(y−2b−a)=0 at the common end of their latus rectum, is