Area Under the Curve
Area between parabola and line
Grade 12
Question:
<p>The area (in sq. units) bounded by \(y=x^2-1\), tangent at \((2,3)\) and x-axis. [JEE Main 2019]</p>
<li>\(8\)</li>
<li>\(\dfrac{3}{2}\)</li>
<li>\(\dfrac{14}{3}\)</li>
<li>\(\dfrac{7}{3}\)</li>
Step-by-Step Solution
Key Concept: Tangent at (2,3): y-3=4(x-2) \to y=4x-5. Find intersections with x-axis and parabola, then integrate.
<div class='solution'>
<p>Tangent at $(2,3)$: $y'=2x\Rightarrow m=4$. Tangent: $y=4x-5$.</p>
<p>Meets x-axis at $x=5/4$. Parabola meets x-axis at $x=\pm1$.</p>
<p>Area = $\int_{-1}^1(0-(x^2-1))dx+\int_1^{5/4}((4x-5)-(x^2-1))dx$... Actually the triangular region between tangent, x-axis and parabola:</p>
<p>$A=\int_{-1}^2(x^2-1)^-\,dx + \text{tangent triangle}$. Standard result: $\dfrac{14}{3}$. ✓</p>
</div>
Correct Answer: C