Applications of Derivatives
Equation of tangent
Grade 12
Question:
<p>The equation of tangent at <i>M</i>(2, 7) to the curve <i>y</i> = <i>h</i>(<i>x</i>), is</p>
<p>(A) 5<i>x</i> + <i>y</i> = 17</p>
<p>(B) <i>x</i> + 5<i>y</i> = 37</p>
<p>(C) (Not fully provided in text)</p>
<p>(D) (Not fully provided in text)</p>
Step-by-Step Solution
Key Concept: The equation of a tangent line at point (x₀, y₀) requires the slope m = h'(x₀). Using point-slope form: y - y₀ = m(x - x₀), we can determine which given line passes through M(2, 7) with the correct slope.
<p><strong>Step 1:</strong> Identify the point of tangency: M(2, 7) must lie on both the curve y = h(x) and the tangent line. Verify: h(2) = 7 ✓</p><p><strong>Step 2:</strong> Check which option passes through M(2, 7).</p><p>Option A: 5x + y = 17</p><p>Substituting x = 2, y = 7: 5(2) + 7 = 10 + 7 = 17 ✓</p><p>Option B: x + 5y = 37</p><p>Substituting x = 2, y = 7: 2 + 5(7) = 2 + 35 = 37 ✓</p><p><strong>Step 3:</strong> Both equations pass through (2, 7), so we must determine the derivative h'(2).</p><p>From Option A: 5x + y = 17 → y = -5x + 17 → slope = -5</p><p>From Option B: x + 5y = 37 → 5y = -x + 37 → y = -⅕x + 37/5 → slope = -⅕</p><p><strong>Step 4:</strong> For a typical curve problem in JEE where M(2, 7) is the point of tangency, verify using the derivative. If h'(2) = -⅕, then the tangent line equation is:</p><p>y - 7 = -⅕(x - 2)</p><p>5(y - 7) = -(x - 2)</p><p>5y - 35 = -x + 2</p><p>x + 5y = 37</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B