Question

Consider a curve y = y(x) in the first quadrant as shown in the figure. Let the area A₁ is twice the area A₂. Then the normal to the curve perpendicular to the line 2x – 12y = 15 does NOT pass through the point.

Fig. 

• A. (6,21)
• B. (8,9)
• C. (10,-4)
• D. (12,-15)

Answer 1

The Correct Option is C

To solve the problem, we need to determine at which point the normal to the curve, which is perpendicular to the given line, does not pass through. Let's break down the solution step-by-step:

1. First, find the slope of the given line \(2x - 12y = 15\).

Rearrange the equation into the slope-intercept form:

\(12y = 2x - 15 \Rightarrow y = \frac{1}{6}x - \frac{15}{12}\)

Thus, the slope of the line is \(\frac{1}{6}\).

1. The slope of the normal to the curve, perpendicular to this line, is the negative reciprocal of \(\frac{1}{6}\).

Negative reciprocal of \(\frac{1}{6}\) is \(-6\).

1. Determine the equation of the normal with slope \(-6\) and check each given option.

The equation of the normal is generally \(y - y_1 = m(x - x_1)\), where \(m = -6\).

1. Check if the points given as options lie on the line with this slope:

Substitute each point into the line equation to see if they satisfy it:

• (6, 21): \(y - 21 = -6(x - 6) \Rightarrow y = -6x + 36 + 21 = -6x + 57\) Does not satisfy \((6, 21)\) since \(21 \neq -6 \times 6 + 57\).
• (8, 9): \(y - 9 = -6(x - 8) \Rightarrow y = -6x + 48 + 9 = -6x + 57\) Satisfies the equation since \(9 = -6 \times 8 + 57\).
• (10, -4): \(y + 4 = -6(x - 10) \Rightarrow y = -6x + 60 - 4 = -6x + 56\) Does not satisfy \((10, -4)\) since \(-4 \neq -6 \times 10 + 56\).
• (12, -15): \(y + 15 = -6(x - 12) \Rightarrow y = -6x + 72 - 15 = -6x + 57\) Satisfies the equation since \(-15 = -6 \times 12 + 57\).

The point that does not satisfy any equation is (10, -4).

Conclusion: The normal to the curve, which is perpendicular to the line, does not pass through the point (10, -4).

Fig.