Question

If \[ P\equiv y^2+2y-x+6=0 \] is a parabola. \(S=0\) is another parabola such that the axes of \(S=0\) and \(P=0\) are same and the distance between their foci is \[ \frac54. \] If the foci of these parabolas are on either side of the vertex of \(S=0\) and the vertex of \(S=0\) is on the directrix of \(P=0\), then the length of latus rectum of \(S=0\) is

• A. \(1\)
• B. \(3\)
• C. \(\dfrac{18}{4}\)
• D. \(4\)

Hint:

For the parabola \[ (y-k)^2=4a(x-h), \] remember: \[ \boxed{\text{Focus }=(h+a,k),\qquad \text{Directrix }x=h-a,\qquad \text{Latus rectum}=4a.} \]

Answer 1

The Correct Option is B

Step 1: Write the given parabola in standard form. Given, \[ y^2+2y-x+6=0. \] Completing the square, \[ (y+1)^2=x-5. \] Comparing with \[ (y-k)^2=4a(x-h), \] we get \[ 4a=1,\qquad a=\frac14. \] Hence, Vertex: \[ V_1=(5,-1), \] Focus: \[ F_1=\left(5+\frac14,-1\right)=\left(\frac{21}{4},-1\right), \] Directrix: \[ x=5-\frac14=\frac{19}{4}. \]

Step 2: Find the vertex of the second parabola. The vertex of \(S=0\) lies on the directrix of \(P=0\). Hence, \[ V_2=\left(\frac{19}{4},-1\right). \] Since both parabolas have the same axis, \[ S\equiv (y+1)^2=4b\left(x-\frac{19}{4}\right). \] Its focus is \[ F_2=\left(\frac{19}{4}+b,-1\right). \]

Step 3: Use the distance between the foci. The distance between the foci is \[ \frac54. \] Also, the foci are on opposite sides of the vertex of \(S=0\). Hence, \[ \left(\frac{21}{4}-\frac{19}{4}\right)+b=\frac54, \] \[ \frac12+b=\frac54, \] \[ b=\frac34. \] The length of the latus rectum is \[ 4b=4\left(\frac34\right)=3. \] Therefore, \[ \boxed{3}. \] Thus, \[ \boxed{(B)} \] is the correct answer.