Exams › JEE Advanced › Maths

The hyperbola x²/a² - y²/9 = 1 passes through the point (8, 3*sqrt(3)). The normal to the hyperbola drawn at this point passes through which of the following points?

1. (15, -2*sqrt(3))
2. (9, 2*sqrt(3))
3. (-1, 9*sqrt(3))
4. (-1, 6*sqrt(3))

Correct answer: (-1, 9*sqrt(3))

Solution

Substituting gives a² = 16; the normal at (8, 3*sqrt(3)) is 2x + sqrt(3)*y = 25, which passes through (-1, 9*sqrt(3)).

Related JEE Advanced Maths questions

• For the circle represented by (x + c)² + y² = a² and the ellipse given as (x - h)² / b² + y² / a² = 1 (where a, b, c, and h are all positive), if they share a tangent that is horizontal, which condition must be satisfied?
• The expression √(x² + (y - 1)²) - √(x² + (y + 1)²) = K describes a hyperbola when:
• When the circle defined by x² + y² = 1 intersects the rectangular hyperbola xy = 1 at four points (x_i, y_i) for i = 1, 2, 3, 4, which of the following is true?
• The line 3x + 4y = 24 meets the x-axis and y-axis at points A and B, respectively, while the line 4x + 3y = 24 intersects at points C and D. The four points A, B, C, and D are located on which type of curve?
• The curve y = f(x), representing a parabola, is tangent to the line y = x at the point where x = 1. Which of the following is true?
• Given (x, y) ∈ R, where x² + y² = 16, we know y = ±√(16 − x²). When x = 0, y equals ±4. For x = ±4, y equals 0. It is observed that no other integer pairs (x, y) satisfy x² + y² = 16. The set R is defined as {(0, 4), (0, −4), (4, 0), (−4, 0)}. What is one value in the domain of R?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →