Question

\(A(-2, 9)\) and \(B(1, 6)\) are two points on the curve \(y = x^2 + 5\). The coordinates of the point \(C\) on the curve such that the tangent drawn at \(A\) is parallel to the chord \(BC\) is:

• A. \((-5, 30)\)
• B. \((0, 5)\)
• C. \((-9, 86)\)
• D. \((6, 41)\)

Hint:

For a curve \( y = f(x) \), the tangent at a point has slope \( f'(x) \). To find a point where the chord to another point has the same slope, set the chord’s slope equal to the tangent’s slope and solve.

Answer 1

The Correct Option is A

Step 1: Verify points \(A\) and \(B\) lie on the curve.
The curve is \( y = x^2 + 5 \).
For \( A(-2, 9) \): \( y = (-2)^2 + 5 = 4 + 5 = 9 \), which matches.
For \( B(1, 6) \): \( y = (1)^2 + 5 = 1 + 5 = 6 \), which matches.
Both points are on the curve.
Step 2: Find the slope of the tangent at \(A(-2, 9)\).
The curve is \( y = x^2 + 5 \). The derivative (slope of the tangent) is: \[ \frac{dy}{dx} = 2x \] At \( x = -2 \): \[ \text{Slope at } A = 2(-2) = -4 \]
Step 3: Determine the slope of chord \(BC\) and set it equal to the tangent’s slope.
Let point \( C \) be \((x_c, y_c)\). Since \( C \) is on the curve, \( y_c = x_c^2 + 5 \), so \( C = (x_c, x_c^2 + 5) \).
The slope of chord \( BC \) between \( B(1, 6) \) and \( C(x_c, x_c^2 + 5) \) is: \[ \text{Slope of } BC = \frac{(x_c^2 + 5) - 6}{x_c - 1} = \frac{x_c^2 - 1}{x_c - 1} = x_c + 1 \quad (x_c \neq 1) \] The tangent at \( A \) has slope \(-4\), and this must equal the slope of \( BC \): \[ x_c + 1 = -4 \implies x_c = -5 \] Thus, \( y_c = (-5)^2 + 5 = 25 + 5 = 30 \). So, \( C = (-5, 30) \).
Step 4: Verify \( C \) lies on the curve and check options.
Point \( C(-5, 30) \): \( y = (-5)^2 + 5 = 30 \), which matches. This corresponds to option (1).
Check other options:
\( (0, 5) \): \( y = 0^2 + 5 = 5 \), on the curve, slope of \( BC \): \( \frac{5 - 6}{0 - 1} = 1 \neq -4 \).
\( (6, 41) \): \( y = 6^2 + 5 = 41 \), on the curve, slope: \( \frac{41 - 6}{6 - 1} = 7 \neq -4 \).
Option (1) is correct. Final Answer: \[ \boxed{1} \]