Conic Sections के अभ्यास 04
Answer: Step 1: The equation of a hyperbola centered at the origin with foci (±5,0) is given by:
x^2/a^2 - y^2/b^2 = 1
Step 2: The length of the transverse axis is 8, which means the distance between the vertices is 8.
Step 3: The distance between the foci is 2ae, where a is the distance between the center and the vertex, and e is the eccentricity.
Step 4: Substitute the values of the foci and the transverse axis length into the equation:
2ae = 8
Step 5: Since the foci are (±5,0), the distance between the center and the vertex (a) is 5.
Step 6: Substitute the values of a and e into the equation:
2(5)e = 8
Step 7: Solve for e:
10e = 8 e = 8/10 e = 4/5
Step 8: Substitute the value of e into the equation of the hyperbola:
x^2/25 - y^2/b^2 = 1
Step 9: Simplify the equation to get:
x^2 - 25y^2/b^2 = 25
Step 10: The equation of the hyperbola satisfying the given conditions is:
x^2 - 25y^2/b^2 = 25
((x-4)^2)/a^2 - (y-0)^2/b^2 = 1
The equation of the hyperbola satisfying the given conditions is:
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
Where, a^2 = 3^2 - 2^2 = 5 b^2 = a^2 - c^2 = 5 - 1 = 4
Therefore, the equation of the hyperbola is: \frac{x^2}{5}-\frac{y^2}{4}=1
स्टेप 1: फोकस और ऊर्ध्वाधर के बीच की दूरी की गणना करें।
दूरी = |3 - 2| = 1
स्टेप 2: अक्ष के ट्रांसवर्स अक्ष की लंबाई की गणना करें।
ट्रांसवर्स अक्ष = 2 * दूरी = 2 * 1 = 2
स्टेप 3: संयुक्त बीजक द्वारा की जा रही अक्ष की लंबाई की गणना करें।
संयुक्त बीजक अक्ष = 2 * दूरी = 2 * 1 = 2
स्टेप 4: हाइपरबोला केंद्र के संयंत्रों की गणना करें।
केंद्र = (2 + 3) / 2 = 2.5, 0
स्टेप 5: हाइपरबोला के मानक समीकरण में मानों को स्थानांतरित करें।
हाइपरबोला का समीकरण: (x - 2.5)^2 / 2^2 - (y - 0)^2 / 2^2 = 1
BETA
