Question

Find the equation of parabola which has the extremities of latus rectum as $(1,2)$ and $(1,-4)$.

• A. $(y+1{)}^2=3(2x-5)$
• B. $(y+1{)}^2=(2x-5)$
• C. $(y+1{)}^2=-3(2x-5)$
• D. $(y+1{)}^2=-(2x-5)$

Answer 1

The correct option is C $(y+1{)}^2=-3(2x-5)$

Latus rectum is parallel to Y-axis.

So, the axis of the parabola is parallel to X-axis.

Equation of parabola having axis parallel to X-axis are $(y-k{)}^2=4a(x-h)$ or $(y-k{)}^2=-4a(x-h)$

Length of latus rectum $=2-(-4)=6$

$\therefore 4a=6$

Co-ordinates of latus rectum are $(1,2)$ and $(1,-4)$

Therefore, co-ordinates of focus are $(\frac{1+1}{2},\frac{2-4}{2})=(1,-1)$

Therefore, co-ordinates of focus are $(h,k)=(1-a,-1)=(-\frac{1}{2},-1)$ for $(y-k{)}^2=4a(x-h)$

and co-ordinates of focus are $(h,k)=(1+a,-1)=(\frac{5}{2},-1)$ for $(y-k{)}^2=-4a(x-h)$

So, equation of parabola are $(y+1{)}^2=6(x+\frac{1}{2})$ or $(y+1{)}^2=-6(x-\frac{5}{2})$

So, the answer is option (C).