Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (-5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point (α,25) on the hyperbola is p, then 4p is equal to:
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Step-by-Step Solution
Step 1: Determine the center and orientation of the hyperbola
The given focus is F1=(−5,0) and the corresponding directrix is x=−59. Since the y -coordinate of the focus is 0 and the directrix is a vertical line, the transverse axis of the hyperbola lies along the x -axis. The center of the hyperbola is (h,k)=(0,0) because the focus is (−c,0) and the directrix is x=−a/e, which implies c=5 and a/e=9/5.
Step 2: Calculate eccentricity and semi-transverse axis
For a hyperbola with center at the origin and transverse axis along the x -axis, the focus is at (−c,0) and the directrix is x=−a/e. From the given information, we have c=5 and a/e=9/5. We can find the eccentricity e by dividing c by a/e, which gives e=5/(9/5)=25/9. Then, we can find a using a=c/e=5/(25/9)=9/5.
Step 3: Calculate semi-conjugate axis
The relationship between a, b, and e for a hyperbola is b2=a2(e2−1). Substituting the values of a=9/5 and e=25/9, we calculate b2. This gives b2=(9/5)2((25/9)2−1)=(81/25)(625/81−1)=(81/25)(544/81)=544/25.
Step 4: Find the product of focal distances
For a point P(x, y) on a hyperbola, the product of its focal distances is given by PF1⋅PF2=e2x2−a2. We are given the point P(α,25). First, we substitute the coordinates of P into the hyperbola's equation a2x2−b2y2=1 to find α2. Then, we substitute α2, e2, and a2 into the focal distance product formula to find p.
Step 5: Calculate 4p
Finally, we need to calculate 4p. We multiply the value of p we found by 4.