Question

The number of real points at which line \(3 y-2 x=14\) cuts hyperbola \(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{4}=5\) is (1) 0 (2) 1 (3) 2 (4) 4

Short answer

1

Step-by-step solution

- Express y in terms of x from the line equation

First, solve the line equation for y. The given line equation is: \[ 3y - 2x = 14 \]. Solving for y: \[ 3y = 2x + 14 \] \[ y = \frac{2}{3}x + \frac{14}{3} \].

- Substitute y into the hyperbola equation

Now, substitute \( y = \frac{2}{3}x + \frac{14}{3} \) into the hyperbola equation: \[ \frac{(x-1)^2}{9} - \frac{(\frac{2}{3}x + \frac{14}{3} - 2)^2}{4} = 5 \]. Simplify the expression inside the parentheses for y: \[ y - 2 = \frac{2}{3}x + \frac{14}{3} - 2 = \frac{2}{3}x + \frac{14 - 6}{3} = \frac{2}{3}x + \frac{8}{3} \]. The hyperbola equation becomes: \[ \frac{(x-1)^2}{9} - \frac{\left(\frac{2}{3}x + \frac{8}{3}\right)^2}{4} = 5 \].

- Simplify and solve the equation

Next, simplify and solve the equation: First, simplify the term \( \frac{\left(\frac{2}{3}x + \frac{8}{3}\right)^2}{4} \): \[ = \frac{1}{4} \times \left( \frac{4}{9}x^2 + \frac{32}{9}x + \frac{64}{9} \right) \text{ (by expanding the square)} \]. This becomes: \[ = \frac{1}{4} \times \left( \frac{4x^2}{9} + \frac{32x}{9} + \frac{64}{9} \right) = \frac{4x^2 + 32x + 64}{36} = \frac{x^2 + 8x + 16}{9} \]. Now, the hyperbola equation is: \[ \frac{(x-1)^2}{9} - \frac{x^2 + 8x + 16}{9} = 5 \] Combine the terms in the numerator: \[ \frac{(x-1)^2 - (x^2 + 8x + 16)}{9} = 5 \] Simplify the numerator: \[ (x^2 - 2x + 1) - (x^2 + 8x + 16) = x^2 - 2x + 1 - x^2 - 8x - 16 = -10x - 15 \]. The equation becomes: \[ \frac{-10x - 15}{9} = 5 \] Multiply both sides by 9 to clear the fraction: \[ -10x - 15 = 45 \].

- Solve for x and determine the number of intersections

Solve the equation \(-10x - 15 = 45\): \[ -10x = 60 \] \[ x = -6 \]. Substitute x back into \( y = \frac{2}{3}x + \frac{14}{3} \) to find the corresponding y value: \[ y = \frac{2}{3}(-6) + \frac{14}{3} = -4 + \frac{14}{3} = \frac{14-12}{3} = \frac{2}{3} \]. Therefore, the line intersects the hyperbola at one point: \((-6,\frac{2}{3})\).

Key concepts

hyperbola equation

A hyperbola is a type of conic section that appears when a plane cuts through both halves of a double cone. The standard equation of a hyperbola centered at \( (h, k) \) with a horizontal transverse axis is given by:
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \].
In the given problem, the hyperbola is centered at \( (1, 2) \) with the equation:
\[ \frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{4}=5 \].
By comparing this with the standard form, it is clear that this hyperbola has been scaled by a factor of 5.
To understand if and where a line intersects this hyperbola, we need to manipulate both equations and find their common points.
This process involves algebraic manipulation and substitution.

solving algebraic equations

Let’s dive deeper into solving equations algebraically. For this exercise, we first rearrange the line equation to express \( y \) in terms of \( x \):
\[ 3y - 2x = 14 \] so, \[ y = \frac{2}{3}x + \frac{14}{3} \].
Next, we substitute \( y \) in the hyperbola equation:
\[ \frac{(x-1)^2}{9} - \frac{\big(\frac{2}{3}x + \frac{14}{3} - 2\big)^2}{4} = 5 \].
Simplifying inside the parentheses:
\[ \frac{2}{3}x + \frac{14}{3} - 2 = \frac{2}{3}x + \frac{8}{3} \],
we insert this back into the hyperbola equation, further simplifying using algebra.
Expanding and combining like terms in the numerator, we derive a single equation in \( x \).
From here, we solve for \( x \) using basic algebraic techniques.
Substitution is a key strategy in these steps, where an equation is algebraically transformed to isolate variables.

coordinate geometry

Coordinate geometry allows us to find the intersection of curves and lines by solving their equations simultaneously.
Given the line \ (3y - 2x = 14) \ and the hyperbola \ ( \frac{(x−1)^2}{9}−\frac{(y−2)^2}{4}=5) \, we use a coordinate approach.
By expressing \( y \) from the line equation and substituting into the hyperbola equation, we translate a geometrical problem into algebra.
This is useful because it allows us to find precise intersection points \( (x, y) \).
In this problem, after performing substitution and simplification, we solve for \( x \). Plugging \( x \) back into the line's equation to find corresponding \( y \) gives us coordinates of the intersection point.
Such techniques are fundamental in coordinate geometry, enabling clear visualization and accurate solutions.