Home
Class 11
MATHS
Find the equation of hyperbola with foci...

Find the equation of hyperbola with foci at the points `(-3,5)` and (5,5) and length of latus rectum `=2sqrt(8)` units.

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation of the hyperbola with foci at the points \((-3, 5)\) and \((5, 5)\) and a length of latus rectum of \(2\sqrt{8}\) units, we will follow these steps: ### Step 1: Find the center of the hyperbola The center of the hyperbola is the midpoint of the line segment joining the foci. \[ \text{Center} (h, k) = \left( \frac{-3 + 5}{2}, \frac{5 + 5}{2} \right) = \left( \frac{2}{2}, 5 \right) = (1, 5) \] ### Step 2: Calculate the distance between the foci The distance between the foci is given by the formula \(2c\), where \(c\) is the distance from the center to each focus. Using the distance formula: \[ 2c = \sqrt{(-3 - 5)^2 + (5 - 5)^2} = \sqrt{(-8)^2 + 0} = \sqrt{64} = 8 \] Thus, \(c = \frac{8}{2} = 4\). ### Step 3: Find the value of \(a\) using the length of the latus rectum The length of the latus rectum \(L\) is given by the formula: \[ L = \frac{2b^2}{a} \] Given that \(L = 2\sqrt{8}\), we can set up the equation: \[ 2b^2 = 2\sqrt{8} \cdot a \implies b^2 = \sqrt{8} \cdot a \] ### Step 4: Relate \(a\), \(b\), and \(c\) For hyperbolas, we have the relationship: \[ c^2 = a^2 + b^2 \] Substituting \(c = 4\): \[ 16 = a^2 + b^2 \] ### Step 5: Substitute \(b^2\) in terms of \(a\) From the previous step, we substitute \(b^2\): \[ b^2 = \sqrt{8} \cdot a \] Substituting this into the equation \(16 = a^2 + b^2\): \[ 16 = a^2 + \sqrt{8} \cdot a \] Rearranging gives us: \[ a^2 + \sqrt{8}a - 16 = 0 \] ### Step 6: Solve the quadratic equation for \(a\) Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(A = 1\), \(B = \sqrt{8}\), and \(C = -16\). \[ a = \frac{-\sqrt{8} \pm \sqrt{(\sqrt{8})^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \] Calculating the discriminant: \[ (\sqrt{8})^2 = 8 \quad \text{and} \quad -4 \cdot 1 \cdot (-16) = 64 \implies 8 + 64 = 72 \] Thus: \[ a = \frac{-\sqrt{8} \pm \sqrt{72}}{2} \] Calculating \(\sqrt{72} = 6\sqrt{2}\): \[ a = \frac{-\sqrt{8} \pm 6\sqrt{2}}{2} \] Choosing the positive root (since \(a\) must be positive): \[ a = \frac{-\sqrt{8} + 6\sqrt{2}}{2} \] ### Step 7: Calculate \(b^2\) Substituting \(a\) back into \(b^2 = \sqrt{8} \cdot a\). ### Step 8: Write the equation of the hyperbola The standard form of the hyperbola is: \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \] Substituting \(h = 1\), \(k = 5\), \(a^2\), and \(b^2\) into the equation. ### Final Equation After calculating \(a^2\) and \(b^2\), we can write the final equation of the hyperbola. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

find the equation of hyperabola where foci are (0,12) and (0,-12)and the length of the latus rectum is 36

Find the equation of the hyperbola with foci at (pm7, 0) and length of the latus rectum as 9.6 units.

The equation of the hyperbola whose foci are (pm 4, 0) and length of latus rectum is 12 is

The equation of the hyperbola whose foci are (pm5,0) and length of the latus rectum is (9)/(2) is

Find the equation of the ellipse whose foci are at the points S(2, 0) and S^(') (-2, 0), and whose latus rectum is 6.

Find the equation of parabola whose focus is (4,5) and vertex is (3,6). Also find the length of the latus rectum.

Find the equation of the hyperbola whose foci are (pm 3, sqrt5,0) and latus rectum is of length 8.

Find the equation of the hyperbola whose foci are (0,+-4) and latus rectum is 12.

Find the equation of the ellipse in the following case: eccentricity e=2/3 and length of latus rectum =5 .

Find the equation of the hyperbola having : vertices (0, +-3) and foci (0, +-5) .