If a hyperbola passes through the point P(10, 16) and it has vertices at (± 6, 0), then the equation of the normal to it at P is :

Solution

Maths sir · 4 steps

Find b² from P first, then write the normal; all four lines pass through P, so the slope decides.

1. Find the hyperbola. The vertices at (± 6, 0) give a = 6, so the hyperbola is x²/36 - y²/b² = 1. Putting in P gives 100/36 - 256/b² = 1, so 256/b² = 64/36 = 16/9 and b² = 144.

   (a² x)/x₁ + (b² y)/y₁ = a² + b²

2. Put in numbers. This is the normal to x²/a² - y²/b² = 1 at (x₁, y₁). With (10, 16) it becomes 36x/10 + 144y/16 = 180, that is 3.6x + 9y = 180. Multiply by 5/9 to get 2x + 5y = 100.
3. Check the slope. The tangent slope at P is (b² x₁)/(a² y₁) = (144 × 10)/(36 × 16) = 5/2. So the normal slope is -2/5, which matches 2x + 5y = 100.
4. So. The normal at P is 2x + 5y = 100, which is option (a).

The idea to keep. When a curve has one unknown constant, use the given point to find it first. For the normal at (x₁, y₁) to a hyperbola, use a²x/x₁ + b²y/y₁ = a² + b².

Answer from the NTA final answer key. Solution by PrepWiser.