Find the equations of the tangent and normal to the hyperbola x²/a² - y²/b² = 1 at the point (x0, y0)
Solution:
Differentiating x2/a2 - y2/b2 = 1 with respect to x we have:
2x/a2 - 2y/b2 dy = 0
⇒ 2y/b2 dy/dx = 2x/a2
⇒ dy/dx = b2x/a2y
Therefore, the slope of the tangent at (x0, y0) is
dy/dx](x0, y0) = b2x0/a²y0
Hence, the equation of tangent at (x0, y0) is
y - y0 = b2x0 / a2y0 (x - x0)
⇒ a2 yy0 - a2 y02 = b2 xx0 - b2 x02
⇒ b2 xx0 - a2 yy0 - b2x02 - a2 y02 = 0
⇒ xx0 / a2 - yy0/b2 - (x02/a2 - y02/b2) = 0
⇒ xx0 / a2 - yy0/b2 - 1 = 0
⇒ xx0 / a2 - yy0 / b2 = 1
Now, the slope of normal at (x0, y0) is
- 1/slope of the tangent at (x0, y0)
= - a2y0 / b2x0
Hence, the equation of normal at (x0, y0) is
⇒ y - y0 = - a2y0 / b2x0 (x - x0)
⇒ (y - y0) / a2y0 = - (x - x0)/b2x0
⇒ (y - y0) / a2y0 + (x - x0) / b2x0 = 0
NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 24
Find the equations of the tangent and normal to the hyperbola x²/a² - y²/b² = 1 at the point (x0, y0).
Summary:
The equation of the tangent and normal to the hyperbola x2/a2 - y2/b2 = 1 at the point (x0, y0) is (y - y0) / a2y0 + (x - x0) / b2x0 = 0
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